Complex Variable Outline. Richard Koch

Transcription

1 Complex Variable Outline Richard Koch March 6, 206

2 Contents List of Figures 5 Preface 8 2 Preliminaries and Examples 0 2. Review of Complex Numbers Holomorphic Functions Extending Functions from R to C Exponential Functions Trigonometric Functions e z as a Map from C to C Log(z) a b The Cauchy-Riemann Equations and Consequences Differentiation Rules The Product Rule The Chain Rule The Inverse Function Rule Complex Functions as Maps from the Plane to the Plane The Cauchy-Riemann Equations The Converse of the Cauchy-Riemann Equations Linear Approximation of Maps R 2 R The Local Geometry of Holomorphic Maps Integration Definition of the Complex Integral A Useful Inequality Consequences of the Inequality

3 CONTENTS 2 5 Review of Advanced Calculus Differentiation Geometry of Subsets Integration Fundamental Theorem of Calculus Differential Forms Solving gradf = E for f Calculus Version of the Residue Theorem Complex Analysis and Advanced Calculus Cauchy s Theorem Power Series Expansions of Holomorphic Functions Indefinite Complex Integrals Morera s Theorem Filling the Gap Summary Complex Analysis from Topology Curves and Homotopy Holomorphic f Have Local Antiderivatives Proof of Homotopy Cauchy Theorem Chapter 6 Done Rigorously Winding Numbers, Homology, and Cauchy s Theorem Winding Numbers The Inside of a Curve The Homological Cauchy Theorem Isolated Singularities and the Residue Theorem The Identity Theorem Liouville s Theorem Isolated Singularities Newton s Calculus Paper Laurent Series Classification of Isolated Singularities The Residue at an Isolated Singularity The Residue Theorem Fundamental Results Meromorphic Functions The Order of a Zero or Pole The Riemann Sphere

4 CONTENTS The Argument Principle The Open Mapping Theorem The Maximum Principle Uniform Limits of Holomorphic Functions Automorphism Groups 94. The Automorphism Group of C The Automorphism Group of the Riemann Sphere The Automorphism Group of the Open Unit Disk The Unit Disk and Non-Euclidean Geometry Topology on H(U) A Metric on H(U) and C(U) The Key Technical Theorem Compact Subsets of H(U) Riemann Mapping Theorem 4 3. The Riemann Mapping Theorem History of the Riemann Mapping Theorem A Short Sketch of Consequences of the Uniformization Theorem Manifolds Riemannian Manifolds Riemann Surfaces Covering Surface The Uniformization Theorem The Theorems of Mittag-Leffler and Weierstrass The Mittag-Leffler Theorem Infinite Products A Convergence Criterion for Products Infinite Products of Holomorphic Functions The Theorem of Weierstrauss Classical Sum and Product Formulas Beautiful Identities A Product Formula The Gamma Function Three Theorems Feynman and Differentiating Under the Integral Sign The Gamma Function The Gamma Function as an Infinite Product

5 CONTENTS The Two Definitions of Γ(z) Agree The Zeta Function 6 7. Some History More History Dirichlet Riemann The Riemann Zeta Function Analytic Continuation of the Zeta Function The Functional Equation of the Zeta Function Bernoulli Numbers The Value of the Zeta Function on Integers Remarks on the Prime Number Theorem A Crucial Calculation The Intuitive Meaning of φ(n) Zeros of ζ(z) on the Boundary of the Critical Strip Proof of the Prime Number Theorem Consequences of the Riemann Hypothesis Other Results in Riemann s Paper Dirichlet Series Dirichlet s Theorem Characters of a Finite Abelian Group Statement of Dirichlet s Theorem; L Functions Analytic Continuation Proof of Theorem Modulo L(, ϕ) 0 for ϕ L(, ϕ) 0 if ϕ Proof of π(x, a) x φ(b) ln x Final Steps in Proof Summary The Theory of Doubly Periodic Functions 28

6 List of Figures 2. f(z) = z(z ) f(z) = Γ(z) f(z) = ζ(z) e z as a Mapping C C Branches of Log(z) γ (z )z(z + ) Various Integration Paths k-dimensional Sets k-dimensional Sets Defining f Defining f Defining f Proving Green s Theorem Region with Holes Extended Green Theorem Cauchy s Formula Goursat s Argument Homotopic Paths Star Shaped Regions Simply Connected Sets f = dg dz Locally A Primitive for the Homotopy A Complicated Curve A Curve with Zero Integral A Region with Three Boundary Curves

7 LIST OF FIGURES Small Path Adjustment Main Diagram of the Proof A Limit Another Limit Existence of Laurent Series The Residue Theorem Contour Integral One Contour Integral Two Proof of the Residue Theorem Stereographic Projection Proof of Inverse Function Theorem Conformal Maps Near a Zero of df dz Uniform Limits in Real Variables Inversion in a Circle Disk is Conformally Equivalent to Upper Half Plane Escher s Angels and Devils Poincare s Model for Non Euclidean Geometry Parallel Postulate Limiting Triangle with Maximum Area Montel s Theorem is False for Real Variables Simply Connected Set with Complicated Boundary m(f) = area fails Glue map ϕ 2 (ϕ ) Modifying Principal Parts to Force Convergence E 2 (z) and E 3 (z) E 4 (z) Modifying E n (z) to Force Convergence ( 4.5 E zi b i 4 z b i ); z i = 2; b i = π cos πz sin πz 6. Euler-Mascheroni Constant New Path for ζ(z) Two Branches of Log(z) Deforming the ζ contour Limiting Case of Homotopy

8 LIST OF FIGURES Correct Graph of ln x Newman s Path sin(0 ln x Approximation of ψ (x) with 00 Zeros of Zeta Conversion Widge is bounded z x

9 Chapter Preface The following material is an outline of complex variable theory. It omits many proofs, but gives enough informal information that a good student could fill in the proofs on their own. There are no exercises. Please consult other books. Some authors prefer to write mathematics in a logical order in which results cannot be mentioned until the tools are available to prove them. An advantage of this style is that the reader always know where they stand because if a result is stated in the text, its proof is nearby. But there are disadvantages, best illustrated with an example. Almost all treatments of complex analysis extend e x, sin x, cos x, and ln x to functions of a complex variable e z, sin z, cos z, and Log(z). This happens early in the text, providing many examples of holomorphic functions, but the extensions can seem arbitrary, one choice out of many. Much later on, these books prove the crucial fact that if a function of a real variable has a holomorphic extension, it is unique. These notes mention that fact as soon as extensions are defined, and sketch a rough proof to be improved later on. The extension of calculus to complex numbers was first done by Euler. Euler discovered that complex analysis provides simple answers to previously unanswered questions, but his techniques often did not meet modern standards of rigor. On the other hand, his results were essentially always correct. Modern books usually postpone stating Euler s beautiful formulas until they can be rigorously proved, but we will often state them as soon as the essential idea is available. For example, Euler proved the following generalization of Wallis product and used it to calculate n= n 2 : sin z z = n= ) ( z2 π 2 n 2 8

10 CHAPTER. PREFACE 9 Euler obtained this result by factoring sin z in terms of its roots, and the key point of his argument was that sin z has no complex roots. We state the result as soon as we know this fact, although a rigorous proof will not be given until much later. The deeper part of complex analysis depends on integration theory. This theory: Cauchy s theorem, Cauchy s formula, the Residue Theorem, and expansion of holomorphic functions as power series, is an immediate consequence of several variable calculus. But many books ignore this fact and develop the theory from scratch, soetimes because their authors believe that topological portions of several variable calculus, like Green s theorem, aren t proved rigorously. In these notes, this theory is developed twice. The first treatment uses several variable calculus. This is followed by a more rigorous treatment using ideas of Emil Artin and Henri Cartan. The essential ideas for complex analysis are already in the portion using several variable calculus. During the more rigorous treatment, a reader already knows the applications can appreciate the ingenious way that Artin and Cartan avoid the Jordan curve theorem and other tricky points in topology, while giving a completely rigorous proof of Cauchy s theorem.

11 Chapter 2 Preliminaries and Examples 2. Review of Complex Numbers A complex number is a point (a, b) in the plane R 2. Instead of writing (a, b), we write a+ib and add and multiply as usual. If we want to speak of a number in general, we usually write z instead of a + ib. If z = (a, b), the usual norm (a, b) = a 2 + b 2 is denoted z. properties of real absolute values. The conjugate of z = a + ib is z = a ib. Notice that zz = z 2. Nonzero numbers have multiplicative inverses given by z = z zz = z z 2 It has the standard so that dividing by z is the same thing as multiplying by z up to a real factor. In particular, the set of complex numbers forms an extension field R C. It is easy to see that there are just two automorphisms of C fixing R, the identity map and conjugation. In polar coordinates we have (a, b) = (r cos θ, r sin θ) where r is the distance of the point from the origin and θ is the angle of the point from the x-axis. Since r = z, with complex notation this formula becomes z = z (cos θ + i sin θ) No doubt you have seen the formula e iθ = cos θ + i sin θ. We ll soon deduce it again. Thus the above formula becomes z = z e iθ 0

12 CHAPTER 2. PRELIMINARIES AND EXAMPLES A consequence of the previous formula is that e iθ =, a fact we ll use again and again. If z = z e iθ and z 2 = z 2 e iθ 2, we have z z 2 = z z 2 e iθ e iθ 2 = z z 2 e i(θ +θ 2 ) So multiplying complex numbers is performed by multiplying their distances from the origin and adding their angles with the x-axis. It is a remarkable fact that complex numbers unify the theories of exponential and trigonometric functions; we ll see more of this soon. The previous formula is a special case of the power law e a e b = e a+b, but changing the notation slightly makes it a special case of the addition formulas in trigonometry: (cos θ + i sin θ ) (cos θ 2 + i sin θ 2 ) = = (cos θ cos θ 2 sin θ sin θ 2 ) + i(sin θ cos θ 2 + cos θ sin θ 2 ) = cos(θ + θ 2 ) + i sin(θ + θ 2 ) 2.2 Holomorphic Functions Complex variable theory is about complex valued functions of a complex variable. Sometimes we denote this w = f(z). There are many examples, like f(z) = z3 +iz+3 z 4 +. Notice that the standard picture of a mountainous surface over the plane only makes sense if the values of f are real. Otherwise, we need four dimensions for the picture. We will soon see that real-valued f are not interesting. On the other hand, old books often contain pictures of the absolute value f(z) of famous functions. The next page shows three examples. The first is a plot of the absolute value. This function has singularities, known as poles, at z = 0,. of z(z ) The second is a plot of the gamma function, extended to be defined for all complex numbers. In calculus, Γ(x) = 0 t x e t dt provided x > 0. Using integration by parts, we have Γ(x + ) = xγ(x); also Γ() =. It follows that Γ(n) = (n )! for all integers n. If we replace x with a complex z, the expression t z e t still makes sense, as we will discover shortly. The integral still converges if R(z) > 0 and defines a function Γ(z). The extension still satisfies the formula Γ(z + ) = zγ(z) for all such complex z. But then we can extend Γ(z) to be defined for all z as follows. By algebra, Γ(z) = Γ(z+) Γ(z+2) z(z+) z = =.... The second equation is defined for R(z + ) > 0 or R(z) >, the third for

13 CHAPTER 2. PRELIMINARIES AND EXAMPLES 2 Figure 2.: f(z) = z(z ) Figure 2.2: f(z) = Γ(z) Figure 2.3: f(z) = ζ(z)

14 CHAPTER 2. PRELIMINARIES AND EXAMPLES 3 R(z) > 2, etc., and in this way we can extend as far left as we like. Notice that poles then appear at 0,, 2,..., as shown in the graph. The final picture shows ζ(z) = n, which converges for R(z) >. This is known as z the Riemann Zeta function because Riemann proved that it can be extended to be defined everywhere except z =, by an argument we will give later. Since n =, the function has a pole as z =. Riemann proved that this function has infinitely many zeros z 0 satisfying 0 < R(z 0 ) < and conjectured that these zeros all have real part 2. Four of these zeros as shown in the graph. This conjecture is the most famous (and probably most important) unsolved problem in mathematics. We define the derivative of a complex valued function by df dz = lim f(z + h) f(z) h 0 h Here h is allowed to be complex, so z + h can approach z from all possible directions in the plane; we require that all of these limits be equal. This is a powerful requirement, with beautiful consequences not true in the real version of the theory. To compute the derivative of f at z, we need to require that f is defined sufficiently near to z in all directions. Thus the domain of f must be an open set U C. We often need to require that U be connected; for instance, if df dz is identically zero, we d like to conclude that f is constant, and that is only true if U is connected. To save words, we define a domain to be a connected open set in C. Complex variable theory, then, is about a special class of functions called holomorphic functions. There are three possible definitions, corresponding to three approaches to the elementary theory. In the end, the three definitions turn out to be equivalent, and after that is proved, the deeper theory opens up in many directions. Definition A complex-valued function f is holomorphic on a domain U if df dz exists at each point of U. (This is our official definition for these notes.) Definition 2 A complex-valued function f is holomorphic on a domain U if df dz each point of U, and the resulting function df dz is continuous on U. exists at Definition 3 A complex-valued function f is holomorphic on a domain U if f is given near each z 0 U by a convergent power series f(z) = c 0 + c (z z 0 ) + c 2 (z z 0 ) The first of these definitions is the least restrictive, but special techniques are required at one or two spots because we do not yet know that df dz is continuous. These notes will sketch this approach.

15 CHAPTER 2. PRELIMINARIES AND EXAMPLES 4 If we assume that df dz is continuous, then the elementary theory is actually a consequence of several variable calculus and can be developed in an hour or two. We ll sketch this approach. The third definition is the most restrictive. It is easy to prove that a function given by a convergent power series is differentiable with derivative given by the term-by-term derivative of the power series, so the third definition implies the first one. An immediate conseqeunce is that a function given by a power series is infinitely differentiable. But in real variable theory, it is easy to find functions which can be differentiated once, but not twice, and it is also easy to find functions which can be differentiated infinitely often and yet do not have convergent power series. So the third definition is only equivalent to the first in complex variable theory. Weierstrass developed complex variable theory by starting with power series. This is possible, but requires tricky proofs that are avoidable by starting with the first or second definition. For example, if f(z) and g(z) are given by power series, then so is f + g, but it is not clear that fg and f/g are given by power series. On the other hand, if f and g have complex derivatives, so do fg and f/g by the ordinary calculus argument, unmodified. Consider the series + z + z 2 + z = z The series only converges for z <, but the function is defined and differentiable everywhere except z =. That s why our third definition has a complicated condition requiring power series centered at each point of the domain. But then we should really prove a theorem stating that if a power series converges for z < R, then the expansion about each z 0 in this disk converges to f at least near z 0. This is true, but the proof becomes trivial if we start with definition one or two. On the other hand, power series are the most natural way to get complex analogues of e x, sin x, cos x, and ln x, so we start with a few elementary facts about them. 2.3 Extending Functions from R to C We are about to extend e x from a function defined on the reals to a function defined on the complex numbers. It is reasonable to expect that this is possible in many ways, and that our extensions are chosen for historical reasons only. Amazingly, this is false. It turns out that there is at most one way to extend a function defined on a subset of R to a holomorphic extension over the complex numbers.

16 CHAPTER 2. PRELIMINARIES AND EXAMPLES 5 Most standard functions of calculus satisfy identities of various kinds. For example, sin 2 x+ cos 2 x =. Amazingly, these identities remain true over the complex numbers, and even more amazingly, there is an abstract theorem which proves this without bothering to check any special case. Here are the two theorems in question, known as the identity theorems. Theorem Suppose U is a domain such that U R. If f is a complex valued function defined on U R, then f can be extended to a holomorphic function on U in at most one way. Theorem 2 Suppose U is a domain such that U R. Suppose f(z) and g(z) are holomorphic on U and f(x) and g(x) satisfy an algebraic identity on U R. Then f(z) and g(z) satisfy the same identity on all of U. Remark: We have to postpone the proofs until later, but a key idea can be given here. We ll sketch the proof of the first theorem from the power series definition of holomorphic functions. Pick a point z 0 in U R and expand f g about z 0. The coefficients of this power series are f k (z 0 ) g k (z 0 ) k!. Since f g is identically zero on U R, all of these coefficients vanish and the power series is identically zero. So f g is zero in an open disk about z 0. Consider the set V of all points z 0 U such that f g = 0 in an open disk about z 0. This set is trivially open; by the previous paragraph it is not empty. We prove it closed, so by that connectedness of U it is all of U. Suppose z n V and z n z 0 U. Then f k (z n ) g k (z n ) = 0. But f g can be expanded in a power series near z 0, so all derivatives of f g exist and are continuous near z 0. It follows that all of these derivatives vanish at z 0, so the coefficients of the series expansion at z 0 are zero and f g is identically zero near z 0. Thus V is closed. 2.4 Exponential Functions The formal course begins here. We need fundamental theorems about power series. Theorem 3 Consider an arbitrary power series about the origin: c 0 + c z + c 2 z There is a constant R, 0 R, such that the series converges absolutely for z < R and diverges for z > R. This R is called the radius of convergence and its value is lim sup c n n. Exercise: Find an example where R = 0 and the series converges only at z = 0. Find an example where R =. Find an example where R = and the series diverges at all z

17 CHAPTER 2. PRELIMINARIES AND EXAMPLES 6 with z =. Also find a similar example where the series converges at all z with z =. Finally find a similar example where the series sometimes converges and sometimes diverges at points with absolute value one. Theorem 4 If the radius of convergence is R, the series converges absolutely for z < R, so it can be rearranged at will. Moreover, if R 0 < R, the series converges uniformly for z R 0. Theorem 5 Let the radius of convergence of the following series be R: f(z) = c 0 + c z + c 2 z 2 + c 3 z Then the radius of convergence of the formally differentiated series is also R: c + 2c 2 z + 3c 3 z Moreover, the function f has a complex derivative at each point z with z < R, and this derivative is given by the formally differentiated series. Therefore, f has complex derivatives of all orders in the convergence disk, and each is continuous. Remark: We now use this to extend e x to e z : Definition 4 The exponential function is the holomorphic function The derivative of this function is e z. Theorem 6 e z = + z + z2 2! + z3 3! +... e z+w = e z e w Remark: This follows from the identity theorem. The second version of that theorem has the vague words algebraic relation, but we can just use the first version of the theorem as follows. Fix w real. Consider f(z) = e z+w and g(z) = e z e w. These are holomorphic functions of z which agree for real z and therefore for all z. Now fix a complex z and consider f(w) = e z+w and g(w) = e z e w. By the previous result, these agree for real w, and therefore for complex w. QED. We can also give a more direct proof. First we give it formally. e z e w = ( ) z k w l z k w l n! = = k! l! k!l! n! k!l! zk w l = n=0 k+l=n n=0 k+l=n (z + w) n 0 n! = e z+w Thus the equation e z e w = e z+w is really all possible binomial theorems wrapped up together.

18 CHAPTER 2. PRELIMINARIES AND EXAMPLES 7 Exercise Make this rigorous, recalling that inside the convergence disk, power series converge absolutely and absolutely convergent series can be rearranged. Corollary 7 The function e z has no zeroes. Proof: e z e z =. Remark: Euler liked to think of power series as generalized polynomials. If P (z) is a non-constant polynomial, it takes all possible complex values by the fundamental theorem of algebra. We might try to generalize this result to say that a non-constant holomorphic function defined in the entire complex plane takes all possible complex values, but e z shows that this is false. However, we will later prove that the values of such a function are dense in the plane. Around 86, after everyone thought that basic complex variable theory was complete, Picard surprised everyone by proving that a non-constant holomorphic function on the entire complex plane can omit at most one value. 2.5 Trigonometric Functions Let x be real and notice that e ix = + ix + i2 x 2 + i3 x = 2! 3! ) ) ( x2 2! + x4 4! i (x x3 3! + x5 5! +... cos x + i sin x Changing x to x and adding and subtracting gives e ix + e ix 2 = cos x e ix e ix = sin x 2i The identity theorem says that we must then define Definition 5 cos z = eiz + e iz 2 sin z = eiz e iz 2i = z2 2! + z4 4!... = z z3 3! + z5 5!... =

19 CHAPTER 2. PRELIMINARIES AND EXAMPLES 8 Theorem 8 The functions cos z and sin z are holomorphic in the entire complex plane and satisfy the following equations: d dz cos z = sin z; d dz sin z = cos z cos 2 z + sin 2 z = cos(z + w) = cos z cos w sin z sin w sin(z + w) = sin z cos w + cos z sin w Remark: This theorem follows from differentiating the power series term by term, and from the first identity theorem by the argument used for e z+w = e z e w. Other proofs are possible. For instance, the derivative of cos 2 z +sin 2 z is zero, so the function is constant by an argument to be given soon. Evaluate at z = 0 to discover that the constant is one. Theorem 9 The function sin z is zero at 0, ±π, ±2π,..., and nowhere else. The function cos z is zero at π 2, π 2 ± π, π 2 ± 2π,... and nowhere else. Remark: For example, if sin z = 0, then eiz e iz 2i = 0, so e iz = e iz and therefore e 2iz =. Writing z = x + iy, we obtain e 2ix e 2y =. Taking absolute values, e 2y = and so y = 0. Therefore all zeros of sin z are real, and thus known from trigonometry. Remark: Euler began the theory of calculus using complex numbers, particularly in a large book written in 749. His proofs are eliminating, but often not rigorous by modern standards. Cauchy developed the modern version of the theory in a series of papers written in the 820 s. As an example of Euler s style, consider the following argument. We know that every polynomial P (z) can be factored completely over the complex numbers, so P (z) = C(z λ )(z λ 2 )... (z λ n ) where C is a constant. Since sin z is given by a power series, it is a generalized polynomial and this factorization should still hold. So sin z = Cz(z π)(z + π)(z 2π)(z + 2π)... This expression clearly does not converge, but we can fix that. Euler s main worry was that sin z might have other complex roots. Eventually he discovered the argument of the previous theorem, so all roots are real and the formula is reasonable. When a series sum converges, the terms must go to zero. Similarly, when a product converges, the individual terms must go to. (This is not true if convergence is defined naively, but the correct definition, given later, makes it true.)

20 CHAPTER 2. PRELIMINARIES AND EXAMPLES 9 Let s rewrite the previous formula to make the terms close to. We can multiply terms by constants, and subsume the changes by also changing C. Here is the fixed version, with a new C: ( sin z = Cz z ) ( + z ) ( z ) ( + z ) ( z ) ( + z )... π π 2π 2π 3π 3π and so sin z z = C Dividing the series expansion of sin z by z gives ) ) ) ( ( z2 π 2 ( z2 4π 2 z2 9π 2... sin z = z2 z 3! + z4 5!... Comparing the series with the product when z = 0 gives C = and thus ) ) ) sin z = ( ( z2 z π 2 ( z2 4π 2 z2 9π 2... This formula is entirely correct; later we will prove it rigorously. Setting z = π 2 gives 2 π = which is known as Wallis Product. Comparing the z 2 term of the series for sin z z with the z 2 term of the infinite product gives 3! = k 2 π 2 and so. Similarly and k= k= k= k= k 2 = π2 6 k 4 = π4 90 k 6 = π6 945 Note that our formulas for sin z and cos z give cos(ix) = cosh x and i sin(ix) = sinh x. It follows that cosh 2 x sinh 2 x =.

21 CHAPTER 2. PRELIMINARIES AND EXAMPLES e z as a Map from C to C The correct way to think of f(z) geometrically is as a map from a subset of the plane to another subset of the plane. To picture this more accurately, draw a grid (not necessarily rectangular) on the domain and draw the corresponding mapped grid lines on the image. In the next chapter, we will see discover the geometric restrictions that holomorphy places on these transformation. We want to draw such a picture for e z. Notice that e x+iy = e x (cos y + i sin y) Consequently, horizontal lines with varying x and fixed y map to radial lines with varying distance to the origin and fixed angle; vertical lines with fixed x and varying y map to circles with fixed radius and varying angle. The resulting picture is the following: Figure 2.4: e z as a Mapping C C The shaded region on the left maps to the pie shaped shared region on the right; points far to the left map close to the origin, and points far to the right map to points near infinity. The region on the left covers the entire complex plane except the origin, and the map then repeats periodically with period 2πi. 2.7 Log(z) Recall that a complex z can be written in polar coordinates as z = z (cos θ + i sin θ). We call this angle the argument of z, arg(z). Rewriting slightly, z = z e i arg z.

22 CHAPTER 2. PRELIMINARIES AND EXAMPLES 2 We define the complex logarithm, Log(z), to be is inverse to e z. So We conclude that by definition, e Log(z) ln z +i arg(z) = z = e Log(z) = ln z + i arg(z) The only problem is that arg(z) is multiple-valued; we can add any integer multiple of 2π and get another legal value for arg(z). Consequently, Log(z) is multiple valued, and its possible values differ by integer multiples of 2πi. Notice that Log(z) is defined for every complex number except z = 0. The main theorems about holomorphic functions require single-valued functions. So whenever the logarithm occurs, we need to pick a branch of the logarithm. That is, we need to define a continuous single-valued function θ(z) on the domain U whose value is one of the legal values of the angle at each point, and define for that particular choice of θ. Log(z) = ln z + iθ(z) Often in complex variable theory, we restrict the Log to the domain U formed by removing from the plane zero and all negative real numbers, and require that π < arg(z) < π. Then Log(z) is single-valued on this domain. In the next chapter, we will prove that Log(z) is holomorphic on this domain; the reader can rather easily prove that already. This particular restriction on domain defines what is called the principal branch of the logarithm. Many other domains U in the plane omit the origin and allow us to define a continuous value of arg(z) throughout the domain. We can then define a branch of the logarithm on U by Log(z) = ln z + iθ and all such functions turn out to be holomorphic. Each such choice is called a permissible branch of the logarithm. We will later prove that such a branch exists on U if and only if U is simply connected. But often the fancy theory is not needed and it is quite clear how to define arg(z). The next page shows examples of simply connected domains of the complex plane which omit the origin. Describe one possible choice of arg(z)s for each such domain:

23 CHAPTER 2. PRELIMINARIES AND EXAMPLES 22 Figure 2.5: Branches of Log(z) 2.8 a b As in ordinary calculus, we define a b = e b ln a. Similarly if a and b are complex, we define a b = e b Log(a). This is multiple-valued, and we make it single-valued by choosing an appropriate branch of the logarithm for a. Common sense is often enough to make this choice. We give three examples. Example : The Riemann Zeta function is defined as n= n for Rz >. In this case, n is a positive z integer, so clearly n z = e ln n z and we use the ordinary calculus logarithm. Example 2: The Gamma function is defined as 0 t x e t dt for Rz > 0. Again t is a positive real number, so t z = e ln t (z ) and we use the ordinary calculus logarithm. The rest of this example can be skipped, but shows how we can make sure of the multiple values of the logarithm to do interesting things. We require Rz > 0 in the above formula because otherwise the singularity at the origin is not integrable. On the other hand, e t vanishes so rapidly at infinity that the integral converges there for all z. For this reason, it is interesting to compute I(z) = t z e t dt γ where γ is the curve shown at the top of the next page. Assume the two portions of the path on the right side are actually on the positive x-axis. Of course this uses the integration theory from a later chapter, but let us be courageous and proceed.

24 CHAPTER 2. PRELIMINARIES AND EXAMPLES 23 Figure 2.6: γ The term t z = e Log(t) (z ) and we assume Log(t) = ln t as we integrate along the first portion of the path. Since the limits of integration are at infinity where the integrand goes to zero, this integral exists for all values of z. As we later prove, this and similar integrals define holomorphic functions of z. So I(z) is globally defined. If Rz > 0, the integrand is small near the origin and we can ignore the integral around the circle there. But the imaginary part of Log(z) will increase from 0 to 2π around this circle. Then as we integrate the last portion in the positive direction along the x-axis, we will be integrating e (ln t+2πi)(z ) dt. The real integrals along the positive x-axis define Γ(z), so we conclude that for Rz > 0, ( I(z) = + e 2πi(z )) Γ(z) or I(z) = 2ie πiz ( e πiz e πiz 2i ) Γ(z) = 2ie πiz sin(πz)γ(z) Thus for Rz > 0, we have Γ(z) = 2i e πiz sin(πz) I(z) On the other hand, I(z) is holomorphic for all z, so this formula gives a second holomorphic continuation of the Gamma function to the entire complex plane. We already know this by an easier argument from the identity Γ(z + ) = zγ(z). It turns out that a similar argument analytically continues the zeta function to the entire complex plane, and in that case the second argument is not available to us. Since sin(πz) vanishes at integers, the Gamma function has poles at the non-positive integers. There are no positive poles because I(z) = 0 at positive integers.

25 CHAPTER 2. PRELIMINARIES AND EXAMPLES 24 Example 3: The remaining examples are more complicated and can be skipped. Notice that z = e /2 Log(z) Although the logarithm has infinitely many values, it is easy to check that z defined this way has only two values, which differ only in sign. If we use the principal branch of the logarithm, we discover that z is holomorphic, defined everywhere except the negative x-axis. If x > 0 for x > 0, then this branch of z has values which are positive imaginary numbers just above the negative x-axis and negative imaginary numbers just below the negative x-axis. Another way to think of z is as a double-valued function which goes through all of its values as we wrap around the origin twice. In this case, it is convenient to put the branch line along the positive x-axis. If we start at a positive x and wind around the origin counterclockwise once, the values of z move from x to x. As we wind around again, the values move from x back to x. Example 4: The length of an ellipse is given by an integral which whose integrand equals the square root of a cubic or quartic, depending on the choice of coordinates. If we want to investigate this integral over the complex numbers, we need to study expressions of the form z 3 + az 2 + bz + c. To be concrete, consider the special case z 3 z = (z )z(z + ) The square root is not differentiable at the origin, so we must remove, 0, and from the plane. Earlier, we asserted that Log(z) has a single-valued branch on any simply connected domain. There is actually a more astonishing theorem: if f(z) is holomorphic and nonzero on a simply connected domain U, then Log(f(z)) has a single-valued branch on U. We will prove this later. It is easy to find examples where U is simply connected, but the range f(u) is not simply connected, so we cannot start by finding a branch of the logarithm on the range of f.

26 CHAPTER 2. PRELIMINARIES AND EXAMPLES 25 At any rate, let us make a branch cut from to along the x-axis. The remaining set is simply-connected, and thus we can define Log(z )z(z + ) on this set. Then (z )z(z + ) = e 2 Log(z )z(z+) is a branch on U. See the left side of the Figure below. Figure 2.7: (z )z(z + ) Notice that (z )z(z + ) = z z z +. If we traverse a very small circle centered at, z and z + hardly change and z z + return to their original value. But z changes sign, and thus (z )z(z + ) changes sign. A similar argument applies if we traverse a small circle about 0. But z already changes sign on the right side, so it does not change sign on the left side. Consequently the entire square root does not change sign on [, 0] and we can remove that portion of the branch cut. We conclude that our branch of (z )z(z + ) is defined on the entire plane except the intervals (, ] [0, ]. The function consists of two copies with this domain and differing only in sign. If we cross one of the branch cuts (9, ] [0, ], we must jump to the other copy of our function, i.e., change the sign of the integrand.

27 Chapter 3 The Cauchy-Riemann Equations and Consequences The Cauchy-Riemann equations were discovered by Cauchy around 820. They were later used extensively by Riemann in his thesis and subsequent paper on Riemann surfaces. Hence the equations are named after both men. 3. Differentiation Rules Theorem 0 The standard differentiation techniques work for df dz. These include the rules for sums, products, and quotients, and the chain rule. Remark: Most proofs work exactly as before. The chain rule requires slightly more work. 3.. The Product Rule 3..2 The Chain Rule 3..3 The Inverse Function Rule 3.2 Complex Functions as Maps from the Plane to the Plane Advanced calculus is partly about maps R m R n. In the special case m = n = 2, such a map is often written (x, y) (u, v) = (u(x, y), v(x, y)). Each function of a complex 26

28 CHAPTER 3. THE CAUCHY-RIEMANN EQUATIONS AND CONSEQUENCES 27 variable defines such a map. Rather than writing u and v, we almost always call the components f and f 2. So f(z) : (x, y) (u, v) = (f (x, y), f 2 (x, y)). Example : Since e z = e x+iy = e x (cos y +i sin y), we have f (x, y) = e x cos y and f 2 (x, y) = e x sin y. Example 2: Since Log(z) = z +iθ, we have f (x, y) = x 2 + y 2 and f 2 (x, y) = arctan y x. Example 3: If f(z) = z 2, we have f (x, y) = x 2 y 2 and f 2 (x, y) = 2xy. Example 4: If f(z) = sin z = sin(x, y) = sin x cos iy+cos x sin iy = sin x cosh y i cos x sinh y. So f (x, y) = sin x cosh y and f 2 (x, y) = cos x sinh y. 3.3 The Cauchy-Riemann Equations Theorem Suppose f(z) = f (x, y) + if 2 (x, y). If df dz exists at a point, then the four partial derivatives f x, f y, f 2 x, and f 2 y all exist at the point and f x = f 2 y f y = f 2 x Proof: Recall that df dz requires that limits be equal as we approach the limit from any direction. In particular, the horizontal and vertical limits must be equal. So df dz = lim (f (x + h, y) + if 2 (x + h, y)) (f (x, y) + if 2 (x, y)) h 0 h f (x + h, h) f (x, y) f 2 (x + h, y) f 2 (x, y) = lim + i lim h ) h h 0 h = f x + i f 2 x The vertical limit is formed by replacing h by ih for a real h: df dz = lim (f (x, y + h) + if 2 (x, y + h)) (f (x, y) + if 2 (x, y)) h 0 ih f (x, h + h) f (x, y) f 2 (x, y + h) f 2 (x, y) = lim + i lim h ) ih h 0 ih = f 2 y i f y

29 CHAPTER 3. THE CAUCHY-RIEMANN EQUATIONS AND CONSEQUENCES 28 Example: If f(z) = e z = e x cos y + ie x sin y then f x = ex cos y = f 2 y f 2 y = ex sin y = f 2 x Show that a real-valued holomorphic function f(z) on a domain U is con- Exercise: stant. 3.4 The Converse of the Cauchy-Riemann Equations Theorem 2 Let f(z) = f (x, y)+if 2 (x, y) on a domain U. Suppose f x, f y, f 2 x, f 2 y exist and are continuous on U, and suppose f and f 2 satisfy the Cauchy-Riemann equations on U. Then df dz exists at each point of U and is continuous on U. Remark: Notice that this is not an exact converse of the previous theorem. We will eventually fill the gap between these theorems by proving that is df dz exists at each point of U, then this derivative is continuous on U and thus all four partials are continuous on U. Sketch of Proof: We want to apply the mean value theorem, which asserts that there is an intermediate value θ between x and x + h with We ll use this to approximate F (x + h) F (x) = h df dx (θ) f(x + h, y + k) f(x, y) h + ik The idea is to move from (x + h, y + k) to (x, y) on variable at a time, and apply the mean value theorem to each move. If we move f (x + h, y + k) to f (x + h, y) and then move f (x + h, y) to f (x, y), we obtain k f y (x + h, θ ) + h f x (θ 2, y) If we move if 2 (x + h, y + k) to if 2 (x + h, y) and then move if 2 (x + h, y) to f 2 (x, y), we get ik f 2 y (x + h, θ 3) + ih f 2 x (θ 4.y)

30 CHAPTER 3. THE CAUCHY-RIEMANN EQUATIONS AND CONSEQUENCES 29 If each of these expressions were evaluated at x, y, we could apply the Cauchy-Riemann equations to obtain h + ik f ih k f 2 (x, y) + (x, y) h + ik x h + ik x which would equal f x + i f 2 x = df dz So our expression is this last result plus a sum of differences, each difference being a partial evaluated one place minus the same partial evaluated somewhere else. This difference goes to zero in the limit by continuity of the partials, so lim f(x+h,y+k) f(x,y) h+ik exists. Remark: This theorem gives an alternate way to extend functions to the complex plane and prove the extension holomorphic. For example, Log(z) = z +i arg z is holomorphic because f (x, y) = x 2 + y 2 and f 2 (x, y) = arctan y x satisfy the Cauchy-Riemann equations. 3.5 Linear Approximation of Maps R 2 R 2 Next we want to discover the geometric meaning of the Cauchy-Riemann equations. This requires a brief revisit to second year calculus. Theorem 3 Let F (x, y) = (u(x, y), v(x, y)) : U R 2 be a map given by functions with continuous partial derivatives on U. Let p = (a, b) U. Then F has a unique linear approximation near p. Indeed, each linear transformation A : R 2 R 2 defines a map such that ɛ : U R 2 ( x a F (x, y) = F (a, b) + A y b There is exactly one linear transformation A such that lim (x,y) (a,b) ɛ(x, y) (x a, y b) = 0 ) + ɛ(x, y) Proof: First we prove uniqueness. If two matrices A and B both work, then subtraction gives ( ) x a (A B) = ɛ y b 2 (x, y) ɛ (x, y)

31 CHAPTER 3. THE CAUCHY-RIEMANN EQUATIONS AND CONSEQUENCES 30 Then ( x a (A B) y b (x a, y b) Select a positive real λ and multiply the vector (x a, y b) by λ. Notice that λ cancels out of the ( expression ) on the left, and yet as λ 0, λ(x a, y b) 0. We conclude that x a (A B) is identically zero, and thus that A = B. y b Existence is proved using the mean-value theorem techniques introduced earlier. Namely, u(x, y) u(a, b) = u(x, y) u(x, b) + u(x, b) u(a, b) = (y b) u y (x, θ ) + (x a) u x (θ 2, b). This can be rewritten as the expression we really want, plus ɛ, where ) 0 u(x, y) = u(a, b) + u u (a, b)(x a) + (a, b)(y b) + ɛ(x, y) x y ( u ɛ(x, y) = x (θ 2, b) u ) (a, b) x ( u (x a) + y (x, θ ) u (a, b) y ) (y b) and a similar result holds for v. Since partial derivatives are continuous, an easy argument concludes the proof. 3.6 The Local Geometry of Holomorphic Maps Next we apply the previous theorem to a holomorphic map f(z) = (f (x, y), f 2 (x, y)). The theorem states that this map has a unique linear approximation at z 0 = a + ib, given by the matrix evaluated at (a, b). f x f 2 x By the Cauchy-Riemann equations, this equals f y f 2 y f x f 2 x f 2 x f x Note that df dz = f x + i f 2 x = df dz (cos θ + i sin θ)

32 CHAPTER 3. THE CAUCHY-RIEMANN EQUATIONS AND CONSEQUENCES 3 where θ = arg ( ) df dz. The above matrix is then ( ) df cos θ sin θ dz sin θ cos θ Theorem 4 Suppose f(z) is holomorphic and df dz is continuous. ( ) Then near z 0 = a + ib, the map f : R 2 R 2 magnifies by df dz and rotates by arg df dz. In particular, maps given by holomorphic f are conformal; that is, they preserve angles. Remark: This theorem is not very informative at zeroes of df dz because the linear approximation at such points is zero. We later find more precise geometrical information near such points. Remark: Notice that the determinant of these linear approximations of holomorphic f are greater than or equal to zero. It follows that holomorphic maps preserve orientation and thus cannot be reflections. The reader should directly show, from the Cauchy-Riemann equations and again directly from the definition, that f(z) = z is not holomorphic.

33 Chapter 4 Integration 4. Definition of the Complex Integral We now begin integration theory. It is a remarkable fact that the deep features of complex analysis depend on integration theory. When I was a graduate student, I met a mathematician from Florida who was able to prove these deep results while avoiding integration. Unfortunately, his proofs were much harder than the canonical proofs which follow. If f(z) is continuous on a domain U, we want to define f(z)dz The ordinary calculus integral f(x)dx should appear as a special case of this complex integral, so the complex integral must be a line integral over paths in U rather than a double integral of some sort. Below is a picture of three possible paths. Figure 4.: Various Integration Paths We do not require that f be holomorphic. It is complex valued, but in special cases f may be real and form a sort of curtain over the path. In these cases, it is easy to guess that f(z) dz gives the area of this curtain, but this is false. Leibniz knew what he was 32

34 CHAPTER 4. INTEGRATION 33 doing when he invented the modern notation of integration. The dz is important. The notation tells us that dz will be real when we integrate horizontally, but imaginary when we integrate vertically. We could, of course, define integration so that it always give the area of the curtain, but then the fundamental theorem of calculus would be false and we definitely need that theorem. Definition 6 A parameterized path is a piecewise continuously differentiable path z(t) = x(t) + iy(t) defined on an interval a t b. Remark: Instead of z(t), we often write γ(t). Too many z s lead to confusion. Remark: Thus there is a finite decomposition a = t 0 < t <... < t n = b such that x(t) and y(t) have continuous derivatives on [t i, t i ]. This includes one-sided derivatives at the t i. Notice that the derivatives at the t i from the left and right need not agree, i.e., the path can have corners. Definition 7 Let f(z) be a continuous complex valued function on a domain U, and z(t) be a parameterized path in U. Then b f(z) dz = f(γ(t)) dγ dt dt γ a Example: Let f(z) = z 2 and let γ be the counterclockwise semicircle with radius starting at z = and ending at z =. Thus z(t) = cos t + i sin t, 0 t π. Then we get the following; all details are given because we want to make a point: π z 2 dz = (cos(t) + i sin(t)) 2 ( sin t + i cos t) dt = π π 0 γ π +i 0 π 0 0 ( (cos 2 sin 2 t)( sin t) 2 cos(t) sin(t) cos t ) dt ( 2 cos t sin t sin t + (cos 2 t sin 2 t) cos t ) dt = ( 3 cos 2 t sin t + sin 3 t ) π ( dt + i 3 sin 2 t cos t + cos 3 t ) dt = 0 ( 3 cos 2 t sin t + sin t( cos 2 t) ) π ( dt + i 3 sin 2 t cos t + cos t( sin 2 t) ) dt = 0 π ( 4 cos 2 t sin t + sin t ) π ( dt + i 4 sin 2 t cos t + cos t ) dt = 0 0 ( ) 4 cos3 t π cos t i ( ) 4 sin3 t π + sin t = ( ) + 0 = 2 3

35 CHAPTER 4. INTEGRATION 34 Example Continued: Or γ z2 dz = z3 3 = 3 3 = 2 3. Theorem 5 Suppose the continuous f(z) is the complex derivative of a holomorphic function dg dz. Then f(z) dz = g(end of γ) g(beginning of γ) Proof: γ b γ f(z)dz = a a γ dg dt dt = ( g dx x dt g 2 x dy dt b a ( g x + i g 2 x ) dt + i Apply the Cauchy-Riemann equations to get b ( g dx x dt + g ) dy dt + i y dt b a b a b a ) ( ) dx dt + idy dt = dt ( g2 dx x dt + g x dy dt ) dt ( g2 dx x dt + g ) 2 dy dt = y dt d b dt g d (x(t), y(t))dt+i a dt g 2(x(t), y(t))dt = (g + ig 2 ) b a = g(end of γ) g(beginning of γ) Remark: Notice that the complex integral is just a special case of the line integral studied in advanced calculus. Indeed if (E x, E y ) is a vector field and γ is a piecewise continuously differentiable path, we defined b ( dx E dγ = E(x(t), y(t)) dt, dy ) dt dt Thus our integral is given by two line integrals b a ( ) dx (f + if 2 ) dt + idy dt = dt γ b a a ( dx (f, f 2 ) dt, dy ) dt + i dt b a ( dx (f 2, f ) dt, dy ) dt dt Remark: Summarizing, the complex integral of an f = dg dz is calculated by the fundamental theorem, exactly as in ordinary calculus. The integral can be calculated more generally for any continuous f, and then the calculation requires the calculation of two line integrals from advanced calculus. Consequently, it is not necessary to discuss in detail the many special cases and peripheral matters already discussed in advanced calculus. We simply list them so the reader can check them off as known:

36 CHAPTER 4. INTEGRATION 35 We always assume that the parameter of integration is a single interval [a, b]. But in practice we integrate over, say, rectangles by integrating separately over each side, and do not bother to adjust parameterizations to be portions of a single interval. Thus each side could be parameterized by [0, ] Suppose t [a, b] is actually t(u) for u [c, d], where t(c) = a and t(d) = b. Then by the chain rule, the integral over [a, b] equals the integral over [c, d]. In short, equivalent parameterizations give the same integral. The line from (a, b) to (c, d) can be parameterized as (a + tc, b + td) where 0 t. The circle of radius r centered at (a, b) is (a + r cos t, b + r sin t) where 0 t 2π. The graph of the function y = h(x) can be made into a parameterized path by t (t, h(t)) Important Remark: Let us integrate z n around a complete circle of any radius about the origin. Here the power n can be any integer, positive or negative. Since z n is the derivative of zn+ n+ and since the path begins and ends at the same point, all of these integrals are zero, except one. If n =, it does not make sense to divide by n +, so we must integrate z directly. We will do this calculation two different ways. Parameterize the circle as re it for 0 t 2π. Then 2π γ z dz = 2π 0 re it rieit dt = i dt = 2πi 0 Alternately, γ z dz = γ dlog(z) dz = Log(z) 2π 0 dz Recall that the logarithm is a multiple valued function. At z moves around the circle counterclockwise, the argument of z increases by 2π. Recall that Log(z) = ln z + i arg z. Thus the integral is dz = 2πi z γ Remark: Curiously, the point made in the previous example already occurred in the first public paper Newton wrote about the calculus. Newton integrated by expanding functions in series and integrating term by term. Some of his expansions had both positive and negative powers of x. In one example, Newton was faced with integrating such a series, and he integrated each term except the x term by using the fundamental theorem of calculus. Newton put a box around x, meaning that he would have to integrate that using numerical methods.

37 CHAPTER 4. INTEGRATION 36 The Newton paper was placed in the Royal Society s offices in London, and there Leibniz read the paper sometime before 672. Leibniz notes survive. When he came to the above spot, he wrote x rather than putting a box around x. Remark: At this spot, I recommend taking time to compute several integral, both using the line integral approach and using the fundamental theorem of calculus, just to make sure that the ideas become straightforward and natural. 4.2 A Useful Inequality Theorem 6 Let f(z) be a continuous complex valued function and γ a piecewise continuously differentiable path. Then ) f(z) dz (max f(z) (length of γ) γ γ Proof: Ultimately the integral is b dz a f(z(t)) dt dt. This is an ordinary integral of a (complexvalued) function of a real variable, so γ b f(z) dz f(z(t))dz dt dt max f(z) γ a b a dz dt dt and a brief calculation shows that the remaining integral gives the length of γ. 4.3 Consequences of the Inequality Corollary 7 Let f n (z) and f(z) be continuous complex-valued functions defined on the range of a piece-wise differentiable curve γ(t), a t b. Suppose f n (z) f(z) uniformly on this image. Then γ f n(z) dz γ f(z) dz. Corollary 8 Let f n (z) and f(z) be continuous complex-valued functions defined on the range of a piece-wise differentiable curve γ(t), a t b. Suppose n= f n(z) = f(z) uniformly on this image. Then n= γ f n(z) dz = γ f(z) dz.

38 Chapter 5 Review of Advanced Calculus 5. Differentiation Two dimensional calculus is about functions like f(x, y) and vector fields like (E x (x, y), E y (x, y)). There is a map, grad : {functions} {vectorfields}, given by ( f gradf = x, f ) y and a second map, curl : {vectorfields} {functions}, given by curl E = E y x E x y We usually think of these maps as forming a sequence {functions} grad {vectorfields} curl {f unctions} The composition of these maps is identically zero because 2 f x y = 2 f y x. 5.2 Geometry of Subsets In the plane, there are certain special subsets called signed 0-dimension sets, signed - dimensional sets, and signed 2-dimensional sets. A signed 0-dimensional set is a finite set of points, each with an attached plus or minus sign. A signed -dimensional subset is a finite set of parameterized curves with an arrow on each indicating a direction to trace the 37

39 CHAPTER 5. REVIEW OF ADVANCED CALCULUS 38 curve. A signed 2-dimensional subset is a region R with the usual orientation, x-first and y-second. There are maps, called boundary maps: {signed 0-dimensional sets} {signed -dimensional sets} {signed 2-dimensional sets} Figure 5.: k-dimensional Sets The boundary of a parameterized curve is a set with two points: the end point with a plus sign and the beginning point with a minus sign. The boundary of a region R is the boundary curve of this region, traced with a direction so that R on the left side. This rule for the boundary of a region gives, for instance, the following picture: Figure 5.2: k-dimensional Sets Notice that the composition of these maps is also zero. 5.3 Integration Consider the sequences introduced in the two previous sections: {f unctions} grad {vectorf ields} curl {f unctions} {signed 0-dimensional sets} {signed -dimensional sets} {signed 2-dimensional sets}

40 CHAPTER 5. REVIEW OF ADVANCED CALCULUS 39 Objects on the top line can be integrated over the corresponding sets on the bottom line. For instance, in the first column, ±f(pi ) where the plus or minus indicates the sign attached to p i. The integral in the second column is the sum of line integrals over the paths in question: (E x, E y ) dγ γ The integral in the third column is R f(x, y)dxdy 5.4 Fundamental Theorem of Calculus At each arrow, there is a theorem relating the integral left of the arrow to the integral right of the arrow. The first of these results is the generalized fundamental theorem of calculus: gradf dγ = f(end of γ) f(beginning of γ) The second result is called Green s theorem: γ R ( ) Ey x Ex dxdy = E dγ y γ= R 5.5 Differential Forms This entire theory generalizes to n-dimensions. The main difficulty is defining the objects which play the roles of the functions and vector fields in two dimensions. In the general case, these are called k-forms, where 0 k n. A single general theorem generalizes the Fundamental Theorem, Green s theorem, and their analogues in three dimensions: the Fundamental Theorem, Stoke s Formula, and the Divergence Theorem. This generalization is known as Stoke s formula. But for complex variable theory, we only need the two dimensional theory sketched above.

41 CHAPTER 5. REVIEW OF ADVANCED CALCULUS Solving gradf = E for f In coordinates, the equation gradf = E becomes f x = E x(x, y) f y = E y(x, y) These equations are the 2-dimensional analogue of df dx = g in one variable calculus, where a solution f is called an indefinite integral of g. In several-variable calculus, the following results are proved about this equation: Theorem 9 If E is a continuous vector field defined on a domain U, and f and f 2 are differentiable functions on U satisfying gradf = gradf 2 = g, then there is a constant c with f 2 = f + c. Theorem 20 The equation cannot be solved unless curl E = 0. Theorem 2 If curl E = 0, the equation can be solved locally. That is, every p U has an open neighborhood V U on which a continuously differentiable f exists with grad f = E. Remark: This subset can be taken to be an open rectangle. We sketch the proof. Solve f x = E x, getting f(x, y) = x a E x (t, y) dt + C(y) Then determine C(y) making f y = E y. Be sure to understand where the hypothesis that curl E = 0 is used in the proof, and how this proof requires a rectangle as domain of E. Remark: For a general domain U, and an E with curl E = 0, there may not exist an f globally defined on all of U with grad f = E. For example, let U = R 2 {0} and define E = ( y, x 2 +y 2 x x 2 +y 2 ). Then no f exists on all of U with grad f = E. Indeed, the appropriate f is arg z, but this is not globally defined on U.

42 CHAPTER 5. REVIEW OF ADVANCED CALCULUS 4 Theorem 22 Let E be continuous on the domain U and fix p U. Define a function f on U as follows: for q U, find a piecewise differentiable path γ from p to q and define f(q) = γ E dγ. Suppose this function is independent of the particular path chosen. Or equivalently, suppose γ E dγ = 0 for all closed curves in U. Then f is well-defined and grad f = E. Figure 5.3: Defining f Remark: It is useful to know the rough idea of the proof. To prove that f x = 0, we find a path from p to the right height for q, and follow that by a horizontal path to q. The initial path is independent of q, so its derivative is zero. An easy calculation shows that the derivative of the integral along the horizontal path is E x. A similar argument using a different path gives f y = E y. Figure 5.4: Defining f Theorem 23 (Crucial Result) If U has no holes and E is a continuous vector field on U with curl E = 0 there is a global f on U with grad f = E. Remark: This is proved by showing that the line integral of E does not depend on the path, or equivalently that the line integral of E around closed curves is zero. This follows from Green s theorem since γ E dγ = curl E over the counterclockwise boundary.

43 CHAPTER 5. REVIEW OF ADVANCED CALCULUS 42 Remark: It is at this point that some readers become concerned about rigor. Have we ever defined the inside of a curve? If a curve can twist arbitrarily, what does it mean to say it is counterclockwise? How precisely was Green s theorem proved? Figure 5.5: Defining f Readers with these questions should carefully read the more rigorous treatment of these points two chapters ahead. For now, it is helpful to recall the proof of Green s theorem. We want curl E. This requires computing two integrals: Ey x Ex y dx dy dy dx In the proof, we integrate these pieces in different orders so we can use the the one-variable fact that the integral of a derivative is the original function. The region of integration must be of a special shape so both of these integrals make sense. We then extend to the general case by breaking the region into pieces, and noticing that the double integral of the entire region is the sum of the double integrals over pieces, and the line integral of the entire region is the sum of the line integrals over pieces. The second fact requires noticing that internal paths have opposite orientation and cancel out. See the pictures below. Figure 5.6: Proving Green s Theorem

44 CHAPTER 5. REVIEW OF ADVANCED CALCULUS Calculus Version of the Residue Theorem The previous material suffices for the following chapters. But it is fun to prove a more general result, which will reappear in complex analysis as one of our most powerful tools. Suppose we have a region with finitely many holes, as indicated below. On this region, suppose we have a vector field E with curl E = 0. Because of the holes, this E need not equal the gradient of a function. Figure 5.7: Region with Holes Around each hole, draw a counterclockwise path γ i. Then a more general form of Green s theorem states Theorem 24 (Extended Green Theorem) Suppose γ is a closed path in a region U with n holes, and let γ i be counterclockwise paths around each hole. Then E dγ = (number of times γ goes around ith hole) E dγ i γ γ holes i Proof: See the pictures below. In each picture, the indicated path does not surround any holes, so Green s theorem states the the integral around this path is zero. But the integral around the outside path equals the sum of the integrals around the holes since other parts of the path cancel. Similar pictures show how to handle paths which go around holes a multiple number of times. Figure 5.8: Extended Green Theorem

45 Chapter 6 Complex Analysis as a Chapter in Advanced Calculus 6. Cauchy s Theorem We now begin the deeper theory. In this chapter, a holomorphic function on a domain U is a function with a complex derivative at each point of U, such that the derivative is continuous. We will ultimately prove that the final condition is automatically true if the first condition is true. Theorem 25 (Cauchy s Theorem) Suppose f is holomorphic on a domain U which contains a closed disk D. Then f(z) dz = 0 Proof: The integral of f also equals the sum of two line integrals: f(z) dz = (f, f 2 ) dγ + i (f 2, f ) dγ D D D The easy way to remember this fact is via Leibniz notation, since we have f(z)dz = (f + if 2 )(dx + idy) = (f dx f 2 dy) + i(f 2 dx + f dy). Apply Green s theorem, and notice that Ey x Ex y equals one of the following two expressions: D f 2 x f y and f x f 2 y Both are zero by the Cauchy-Riemann equations. QED 44

46 CHAPTER 6. COMPLEX ANALYSIS AND ADVANCED CALCULUS 45 Remark: Cauchy s theorem is one of our most important results. We later generalize it to state the γ f(z) dz = 0 if f is holomorphic on a domain which includes γ and every point inside γ. Theorem 26 (Cauchy s Formula) Suppose f is holomorphic on a domain U which contains a closed disk D. For every z 0 strictly inside D, f(z) dz = f(z 0 ) 2πi z z 0 D Remark: This astonishing theorem says that the value of f at every point inside the disk is completely determined by its values on the boundary of the disk. Proof: Let g(z) = f(z) f(z 0) z z 0 when z z 0 and df dz (z 0 when z = z 0. Since f is holomorphic at z 0, g is continuous at z 0 and holomorphic on the rest of U. Consider the curve γ shown below; g is holomorphic everywhere inside γ because z 0 is outside. The proof of Cauchy s theorem still works and shows that the integral of g around γ is zero; essential this is Green s theorem applied to the region inside γ. Figure 6.: Cauchy s Formula On the other hand, the integral of g over the small circle is bounded by the maximum of g near the circle times the length of the circle, and as the circle shrinks to zero, this approaches zero. We conclude that the integral of g around the large outside circle is zero and consequently 2πi D f(z) dz = z z 0 2πi D However, the right side of this expression is f(z 0 ) 2πi D Log(z z 0 ) is z z 0, the integral of z z 0 f(z 0 ) z z 0 dz z z 0 dz. Since the derivative of is Log(z z 0 ). The real part of this expression

47 CHAPTER 6. COMPLEX ANALYSIS AND ADVANCED CALCULUS 46 is single valued, namely ln z z 0, so it comes back to itself as we circle z 0. But the imaginary part of Log(z z 0 ) is arg(z z 0 ), and this increases by 2π as we trace the large circle counterclockwise. So the integral on the right side is 2πi f(z 0)(2πi) = f(z 0 ). QED 6.2 Power Series Expansions of Holomorphic Functions Theorem 27 Let f(z) be holomorphic on a domain U. Let z 0 U. Then f has a power series expansion f(z) = c k (z z 0 ) k which converges to f at least in the largest open disk about z 0 inside U. k=0 Corollary 28 If f is holomorphic on U, it has continuous complex derivatives of all orders. Corollary 29 Suppose a power series k c k(z z 0 ) k converges to f on z z 0 < R. Then at each z in this disk, f is given by a power series k=0 c k(z z ) k centered at z which converges to f near z. Proof of Theorem: First a notational point. Many authors prefer to reserve z for the arguments of functions and let ζ be the variable of integration. These authors write Cauchy s formula as f(z) = f(ζ) 2πi ζ z dζ We won t be dogmatic about the point, but the new notation will be convenient in the following proof because it involves ζ, z, and z 0. Let D be a closed disk centered at z 0 such that the entire disk is inside U. Call the counterclockwise boundary of this disk γ. By Cauchy s formula, for every z strictly inside the disk, f(z) = f(ζ) 2πi γ ζ z dζ = f(ζ) 2πi γ (ζ z 0 ) (z z 0 ) dζ = ( ) ( ) f(ζ) 2πi γ ζ z 0 z z dζ 0 ζ z 0 Since ζ is on the boundary of the disk and z is inside the disk, z z 0 ζ z 0 < γ

48 CHAPTER 6. COMPLEX ANALYSIS AND ADVANCED CALCULUS 47 so we can expand the last parenthetical expression as a series converging uniformly in ζ. This gives f(z) = f(ζ) ( ) z k z0 dζ 2πi γ ζ z 0 ζ z 0 We can move the summation sign outside the integral by the last result in chapter 4, so [ ] f(ζ) f(z) = dζ (z z 2πi γ (ζ z 0 ) k+ 0 ) k k=0 This is the desired power series about z 0. If our disk is contained in a larger disk D, and if γ is the boundary of the larger disk, the same formula holds. But power series expansions are unique, since the coefficients must be f (k) (z 0 ) k! by repeated term by term differentiation. So it follows that the original series converges to f in the larger disk, and therefore in the largest open disk about z 0 in U. k 6.3 Indefinite Complex Integrals Theorem 30 Suppose g, h, and f are holomorphic on a domain U and dg U. Then h = g + c for some complex constant c. Proof: The complex derivative of g = g + ig 2 is g x + i g 2 x = g 2 y i g y dz = dh dz = f on by the Cauchy-Riemann equations. So all partials of g agree with corresponding partials of h. It then follows trivially that g and h differ by a constant. Theorem 3 Suppose g is holomorphic on a rectangle. Then there is a holomorphic f on this rectangle with df dz = g. Thus indefinite integrals of holomorphic g exist locally. Proof: We want to find continuously differentiable real-valued functions f and f 2 with gradf = (g, g 2 ) and gradf 2 = (g 2, g ), because then f x = f 2 y and f y = f 2 x, so f = f + if 2 would satisfy the Cauchy-Riemann equations and by the converse of these equations, f would be holomorphic with derivative g. By several variable calculus, these equations can be solved in a rectangle provided curl(g, g 2 ) = 0 and curl(g 2, g ) = 0. Writing these equations out, we need g 2 x = g y and g x = g 2 y, and these equations hold because g is holomorphic.

49 CHAPTER 6. COMPLEX ANALYSIS AND ADVANCED CALCULUS 48 Remark: Indefinite integrals need not exist globally. For example, z is holomorphic on C {0}, but its indefinite integral would have to be Log(z) + c, and this function is multiple-valued. Remark: We will later prove that if g is holomorphic on a simply connected U, then it has an indefinite integral. 6.4 Morera s Theorem Eventually we will prove that if f is holomorphic on a simply-connected U, then the integral of f around any closed curve in U is zero. At the moment, we know this only in special cases, including integrating around a circle contained with its interior in U. Theorem 32 Let f be holomorphic in an open rectangle. Then for any closed curve γ in this rectangle, γ f = 0. Proof: By the previous theorem, there is a g with f = dg dz (the roles of f and g were reversed in the previous section). So the integral of f over any closed curve is g at the end of the curve minus g at the beginning. Since the beginning and end are the same, this integral is zero. Remark: The converse of the previous result is true, and thus if f = 0 for all closed curves, f must be holomorphic. Indeed Theorem 33 (Morera s Theorem) Suppose f is continuous on a domain U. If we have f(z) dz = 0 R for all rectangles contained, with their interior, in U, then f in holomorphic in U. Proof: The integral of (f +if 2 )(dx+idy) over any rectangle is zero, so the real integrals of (f, f 2 ) and (f 2, f ) over any rectangle is zero. The real variable argument on page 35 then shows that if we fix a rectangle contained with its interior in U, then there exist g and g 2 continuously differentiable on this rectangle with gradg = (f, f 2 ) and gradg 2 = (f 2, f ). It follows that g x = g 2 y and g y = g 2 x. So g +ig 2 is holomorphic with derivative f in the fixed rectangle. But we know that holomorphic functions are infinitely differentiable, so f has a continuous derivative at each point in the fixed rectangle. So f is locally holomorphic, hence globally holomorphic.

50 CHAPTER 6. COMPLEX ANALYSIS AND ADVANCED CALCULUS Filling the Gap At last we are able to prove Theorem 34 Suppose f(z) is continuous on a domain Uand df dz exists at each point of U. Then f is holomorphic in U, i.e., this derivative is continuous. Remark: This gap opened up when we proved the Cauchy-Riemann equations assuming only existence of derivatives, but required continuity of the derivative to prove the converse. We have used the converse of the Cauchy-Riemann equations at several points in the above exposition. To fill the gap, we use the previous theorem. Notice that the proof of that theorem didn t use anything about f except that it is continuous and the integral around all rectangles is zero. So it is enough to prove the following, first proved by Goursat: Theorem 35 Let f(z) be continuous on a domain Ucontaining a closed rectangle, and suppose df dz exists at each point on an inside the rectangle. Then R f(z) dz = 0. Proof: Suppose this is not true, and R f(z) dz = A > 0. Divide the rectangle into four pieces as shown below. The sum of the integrals around these subrectangles equals the original integral, so at least one subrectangle must exist whose integral is greater in absolute value than A/4. Subdivide this rectangle similarly. We obtain a sequence of rectangles R R 2 R 3..., such that R n f dz A/4 n. Figure 6.2: Goursat s Argument

51 CHAPTER 6. COMPLEX ANALYSIS AND ADVANCED CALCULUS 50 By a standard argument, there is a point z 0 in the intersection of these closed rectangles. By assumption, the derivative of f exists at z 0. Thus we can write where f(z) = f(z 0 ) + df (z z 0 ) + ɛ(z) dz z0 lim z z 0 ɛ(z) z z 0 = 0 Therefore, for any positive constant B, we obtain ɛ(z) B z z 0 for z sufficiently close to z 0. But f(z 0 ) + df (z z 0 ) is trivially holomorphic and its derivative around R n is zero. z0 Consequently, dz (maximum of ɛ(z) )(length of R n ) ɛ(z)dz > A/4n R n Notice that each time we divide into four subrectangles, the length of each of these rectangles is half of the original length. So if the original length is L, then length of R n = L/2 n. Also, the maximum possible value of z z 0 in the original rectangle is the diagonal D of the rectangle. Each time we subdivide into four pieces, this maximum decreases by 2. We conclude that ɛ BD/2 n. Putting this all together, B(D/2 n )(L/2 n ) ɛ(z) L/2 n > A/4 n and so BD > A. However, A and D are fixed, but B can be chosen as small as we wish by choosen rectangles sufficiently close to z 0, so we have a contradiction.

52 CHAPTER 6. COMPLEX ANALYSIS AND ADVANCED CALCULUS Summary Suppose f(z) is a continuous function on a domain U. We have proved that the following conditions are equivalent: f is holomorphic on U df dz df dz dk f dz k exists at each point of U exists at each point of U and is continuous on U exists at each point of U for all k f can be expanded in a power series centered at any z 0 U which converges at least in the largest open disk about z 0 in U f is locally dg dz on U R f(z)dz = 0 for all rectangles contained, with their interiors, in U f = f + if 2 where f and f 2 have continuous partial derivatives in U and satisfy the Cauchy-Riemann equations

53 Chapter 7 Complex Analysis from a Topological Point of View This is a chapter for doubters, including those extreme doubters who claim that several variable calculus is hopelessly pictorial and non rigorous. We will reprove everything in the previous chapter from scratch. But don t worry; either the previous proof was sufficient and we ll refer you back to it, or else the previous proof can be dramatically simplified using an powerful version of Cauchy s theorem proved next. In this chapter, a function f is holomorphic on U if df dz exists at all points of U; we do not require continuity of this derivative. The new version of Cauchy s theorem allows us to avoid any mention of Green s theorem. We ll show that this can be done without using fancy results like the Jordan curve theorem, and with complete rigor so every reader will henceforth be willing to deform integration paths with abandon. The topologists introduce two tools to study domains in the plane: the homotopy group π (U) and the homology group H (U, Z). Following their lead, we prove a homotopy version of Cauchy s theorem in this chapter. This is followed in the next chapter by a deeper homological version of the theorem. 7. Curves and Homotopy Let U be a domain in the plane. In this chapter, a path is a continuous curve in U. Notice that we do not require differentiability. All of our paths will be parameterized by [0, ], so a path is a continuous map γ : [0, ] U 52

54 CHAPTER 7. COMPLEX ANALYSIS FROM TOPOLOGY 53 We say two closed paths are homotopic if one can be deformed to the other inside U through intermediate closed paths. Analogously, two paths which start at p and end at q are homotopic if one can be deformed to the other inside U through intermediate paths which also start at p and end at q. See the pictures below. Making this rigorous is easy. map Figure 7.: Homotopic Paths A homotopy between paths γ(t) and τ(t) is a continuous h(t, u) : [0, ] [0, ] U such that h(t, 0) = γ(t) and h(t, ) = τ(t). The intermediate paths of the homotopy are obtained by fixing u 0 and writing h(t, u 0 ). If γ and τ are closed, we have a homotopy through closed paths if h(0, u) = h(, u) for all u. If γ and τ both start at p and end at q, we have a homotopy with fixed endpoints if h(0, u) = p and h(, u) = q for all u. Usually it is clear which type of homotopy is being used, and we simply say homotopic. It is relatively easy to prove paths homotopic. The following general argument is often enough: Definition 8 A domain U is star-shaped with respect to p U if whenever q U, the straight line joining q to p is entirely in U. See the pictures which follow. Figure 7.2: Star Shaped Regions

55 CHAPTER 7. COMPLEX ANALYSIS FROM TOPOLOGY 54 Definition 9 A domain U is simply-connected if every closed path in U is homotopic to a constant path, i.e., a point. Theorem 36 A star-shaped region is simply connected, so any closed path is homotopic to a constant. Proof: Suppose U is star-shaped with respect to p. Notice that h(t, u) = γ(t)( u) + p(u) deforms along lines and thus remains in U. This homotopy starts at γ and ends at p. Corollary 37 The following domains are all simply connected the interior of a rectangle the interior of a disk the entire complex plane Remark: Simple connectivity is a topological property, so if two sets U and V are homeomorphic and one is simply connected, the other is also. This is one way to prove rigorously that a set is simply connected. For example, z z 6 maps one-eight of an annulus to three-fourths of an annulus. The first set is star-shaped, hence simply connected. Therefore the second set is simply connected, but not star-shaped. Figure 7.3: Simply Connected Sets

56 CHAPTER 7. COMPLEX ANALYSIS FROM TOPOLOGY 55 Remark: Finally, here is the big theorem. Theorem 38 (Homotopy Form of Cauchy s Theorem) Suppose f is holomorphic on a domain U and γ and τ are either closed curves in U or else curves in U with the same starting and ending points. If γ and τ are homotopic, then f(z)dz = f(z)dz γ Corollary 39 If f is holomorphic on a simply connected domain U, then for any closed γ in U, f(z) dz = 0 γ Remark: It would be unfortunate if we had to restrict attention to piece-wise differentiable homotopies, because we d constantly need to check differentiability. But remarkably, this isn t necessary. Instead, we ll begin the proof by defining γ f(z)dz when γ is merely continuous, and then allow homotopies which are only continuous. Since continuous curves can have infinite length, and can be space-filling, it seems highly unlikely that we ll be able to define γ f(z) dz over them. But there is a catch: IMPORTANT POINT: We previously defined γ f(z) dz when f is merely continuous on U, but γ is a piecewise-continuously differentiable curve. The new definition will work for all continuous curves, but then f(z) must be holomorphic. The two definitions agree when both are defined. It will turn out that defining the integral over continuous curves is the main difficulty. Once we do that, proving Cauchy s theorem is easy because the same technique is involved. Both tasks ultimately depend on the assertion that if f is holomorphic, then locally there is a holomorphic g with dg dz = f, and for completeness we ll reprove that from scratch. τ 7.2 Holomorphic f Have Local Antiderivatives We reprove this key assertion from scratch. Suppose f is holomorphic on U and find an open rectangle S contained with its interior in U. We only assume that df dz exists at each point of S. Turn back to Goursat s theorem in section 6.5, which asserts that the integral of f around any rectangle R in our S is zero. This argument doesn t use Green s theorem and remains valid here. So we ll reuse it without repeating the proof. We can define g on S by integrating from a fixed point p 0 to z along two sides of a rectangle, either horizontally and then vertically, or else vertically and then horizontally. Goursat s

57 CHAPTER 7. COMPLEX ANALYSIS FROM TOPOLOGY 56 theorem guarantees that these integrals are equal. See the picture below. We will prove that f f x = i y = df dz. It follows that g has continuous partial derivatives, namely f, and satisfies the Cauchy-Riemann equations, so g is holomorphic with derivative f and we are done. Figure 7.4: f = dg dz Locally Take, for instance, the path vertically and then horizontally, illustrated previously. Call the entire path γ and the vertical path γ, and parameterize the horizontal path τ(t) = t + iy for x 0 t x. Then x g(x + iy) = f + f(t, y)dt γ x 0 The first integral is independent of x, so The equation g y g x = d dx x x 0 f(t, y) dt = f(x, y) = if(x, y) is proved similarly. QED.

58 CHAPTER 7. COMPLEX ANALYSIS FROM TOPOLOGY Proof of Homotopy Cauchy Theorem This proof comes from an elegant short book on complex analysis by Henri Cartan. We begin by sketching the key idea behind our new definition of γ f(z) dz for continuous γ and holomorphic f. Consider the special case z dz; γ(t) = e4πit : [0, ] C {0} γ It is tempting to integrate using the fundamental theorem of calculus. Thus d dz log(z) = z ; log(z) = ln z + i arg(z) On the unit circle, ln z = 0, and arg(z) increases by 4π as we circle twice. So the integral is 4π. The trouble with this calculation is that log(z) is multiply-valued, so it doesn t make sense to treat it as an ordinary holomorphic function on our domain. On the other hand, it does make sense to consider log(γ(t)) = ϕ(t) as a single valued function on [0, ], given by ϕ(t) = 4πt. This function is continuous, and the key point is that locally it equals g(γ(t)) where g is a branch of the logarithm satisfying dg dz = z and defined on a smaller V in the domain of z. So the central idea of the proof is that multiply-valued g become single-valued ϕ(t) = g(γ(t)). We ll now sketch the full proof of the homotopy version of Cauchy s theorem. An ambitious reader can fill in the details without further help, but we ll offer a little help at the end. Assume f is holomorphic on U and γ is a continuous path in U. A primitive for γ is a continuous complex valued function ϕ on [0, ] which is locally the pullback of a local g on V U with dg dz = f. Lemma Such a primitive exists. Lemma 2 A primitive is unique up to a constant. Lemma 3 Consequently, ϕ() ϕ(0) is well defined independent of the chosen primitive. This invariant number is called γ f(z) dz. Lemma 4 If γ is piecewise-continuously differentiable, the old and new definitions of the integral agree. Definition 0 Let h(t, u) : [0, ] [0, ] U be a homotopy. A primitive for h is a continuous function ϕ on [0, ] [0, ] which is locally the pullback of a continuous g on V U with dg dz = f.

59 CHAPTER 7. COMPLEX ANALYSIS FROM TOPOLOGY 58 Lemma 5 Such a primitive exists. Proof of Cauchy s theorem for a homotopy of closed curves: Let ϕ(t, u) be a primitive of h on [0, ] [0, ]. Then ϕ(t, 0) is a primitive for γ and ϕ(t, ) is a primitive for τ. Thus γ f = ϕ(, 0) ϕ(0, 0) and τ f = ϕ(, ) ϕ(0, ). Moreover, h(0, u) = h(, u) since all curves are closed, so these are the same curve. Therefore ϕ(0, u) and ϕ(, u) differ by a constant, so ϕ(, u) ϕ(0, u) is a constant independent of u. So γ f = τ f. Proof of Cauchy s theorem for a homotopy with fixed endpoints: The proof is the same as the previous proof until the end. All curves start at p and end at q, so h(0, u) = p and h(, u) = q. Since both of these curves are constant, one possible primitive is a constant, so hϕ(, u) = c and ϕ(0, u) = c 0. So ϕ(, u) ϕ(0, u) = c c 0, and thus γ f = τ f. Remark: PRETTY SIMPLE. But maybe we should give a few more details. Let s start with the definition of a primitive of a curve. So Definition A primitive for a continuous curve γ in U and a holomorphic f on U is a continuous complex-valued ϕ on [0, ] which is locally the pullback of g such that dg dz = f. This means that for every t 0 [0, ] there is an open neighborhood W of t 0 [0, ] and an open neighborhood V of γ(t 0 ) in U and a g on V with dg dz = f such that ϕ(t) = g γ(t) on W. Remark: Please take time to understand this in detail. Everything else will then be easy. Sketch of Proof of First Lemma: Since g exists locally on U, we can find an open cover {W α } of [0, ] with a primitive ϕ α on each W α. By compactness, we can find a subdivision 0 = t 0 < t <... < t n = such that each [t i=, t i ] is in some W α. So a primitive ϕ i exists on [t i, t i ]. Since g is only defined up to a constant, we can add constants to the ϕ i at will. Choose these constants to adjust the ϕ so they match at boundary points t i. QED. Sketch of Proof of Second Lemma: Suppose ϕ and ϕ 2 are primitives. Locally ϕ and ϕ 2 are pullbacks of g and g 2, both with derivative f. By shrinking domains, we can assume that g and g 2 are both defined on the same connected open V. Since their derivatives are equal, they differ by a constant. So ϕ 2 ϕ is continuous and locally constant. Since [0, ] is connected, the difference is globally constant. QED. Sketch of Proof of Fourth Lemma: As earlier, subdivide [0, ] using t i so ϕ is the pullback of a single locally defined g on [t i, t i ]. By the fundamental theorem of calculus, the original definition of the integral gives f(z) dz = g(γ(t i )) g(γ(t i ) = ϕ(t i ) ϕ(t i ) on [t i, t i ]. Both the original integral and the new integral are clearly additive when we subdivide the path, so we are done. Sketch of the Fifth Lemma: As with an earlier argument, we can subdivide [0, ] [0, ]

60 CHAPTER 7. COMPLEX ANALYSIS FROM TOPOLOGY 59 into a grid such that each subrectangle of the grid maps to a V on which a local g exists. So we obtain primitives ϕ ij on each subrectangle, which can be modified by adding constants. Starting at top-left, and working across the rows, and then down row by row, adjust these constants so the ϕ ij match on boundaries, to obtain a globally defined ϕ. It is instructive to work through this and convince yourself that these adjustments are possible. Two tricky situations can occur. These are illustrated below. In each case, a new subrectangle is being added to the diagram. The black curve defines a parameterized path with two primitives, namely the primitive defined by the already constructed piece of h and the primitive from the new subrectangle. So these primitive differ by a constant and we can adjust the new primitive for the new subrectangle by adding a constant to make it continuous across the black boundary. Figure 7.5: A Primitive for the Homotopy 7.4 Chapter 6 Done Rigorously We now know enough to reprove everything in chapter 6 rigorously. We have to do this in the correct order.

61 CHAPTER 7. COMPLEX ANALYSIS FROM TOPOLOGY 60 The first theorem in chapter 6 is Cauchy s theorem that D f(z) dz = 0 if f is holomorphic on U containing the disk D. We already know a very substantial generalization, namely that we can replace D by any closed curve in a simply connected region. Moreover, our original proof required continuity of df dz and our current proof doesn t need this hypothesis. The second theorem in chapter 6 is Cauchy s formula f(z) = D ζ z dz if f is holomorphic on U containing the disk D and z is inside this disk. Look back and notice that our original proof used Green s theorem to prove that the integral of f(ζ) f(z) ζ z over the disk is equal to the integral over a small circle about z. The quotient expression is continuous at z but may not be holomorphic there, but the boundary of D and the small circle are homotopic by a homotopy which avoids z. So these integrals are equal. The original proof then showed that the integral around the small circle approaches zero. It follows that the integral of f(ζ) f(z) ζ z around D is zero. From this point, the original proof still works and easily gives Cauchy s formula. Next we proved theorem 27 that a holomorphic function can be expanded in a power series. This proof works unchanged, except that the requirement that f have a continuous derivative is no longer needed. The easy proof of theorem 30 still works. But theorem 3 and theorem 32 can be replaced by a powerful generalization: Theorem 40 Let f be holomorphic on a simply connected U. Then there is a global g on U with dg dz = g. Proof: Fix p U and for each q U, define f(z) = γ f(z) dz. This f does not depend on the path γ because the integral of f around any continuous closed path is zero by the homotopic Cauchy theorem. The argument used in section 7.2 then shows that dg dz = f. This completes our redo of chapter 6. The previous theorem has powerful corollaries, as well: Corollary 4 If U is a simply connected domain which does not contain 0, it is possible to define a branch of the logarithm on U. Proof: Use the theorem to find g on U with dg dz = z. Then d dz ( e g z ) = zeg z eg z 2 = 0 Consequently e g (z) = cz for some constant c 0. Write c = e λ. Then g λ is a branch of the logarithm because e g λ = e g e λ = e λ ze λ = z f(ζ)

62 CHAPTER 7. COMPLEX ANALYSIS FROM TOPOLOGY 6 Corollary 42 If U is a simply connected domain which does not contain 0, it is possible to choose a single-valued branch of z on U. Proof: Select z = e 2 Log(z). Corollary 43 If f(z) is holomorphic and never zero on a simply connected domain U, it is possible to define a branch of Log(f(z)) on U. Proof: It is easy to find examples where f(u) is no longer simply connected, so this does not follow from an earlier corollary. If such a branch exists, its derivative will be f (z) f(z). This function is holomorphic because f has no zeros on U. Find g with dg dz = f f. Then ( ) d e g(z) dz f(z) = ( ) fe g f f e g f f 2 = 0 So this function is a (nonzero) constant e λ and e g(z) = f(z)e λ. Thus and so g(z) λ is a branch of Log(f(z)). e g(z) λ = f(z) Corollary 44 If f(z) is holomorphic and never zero on a simply connected domain U, it is possible to define a branch of f(z) on U. Proof: f(z) = e 2 Log(f(z))

63 Chapter 8 Winding Numbers, Homology, and Cauchy s Theorem 8. Winding Numbers Previously we talked about domains with holes, always accompanied with a picture. We ll finally be able to define this rigorously. The development which follows was introduced into complex variables by Emil Artin, in his lectures on the subject. Definition 2 Fix a point z 0. Let γ be a closed curve which does not contain z 0. The winding number of γ about z 0 is W (γ, z 0 ) = dζ 2πi ζ z 0 Intuitively, this expression counts the number of times γ goes around z 0 counterclockwise. Clockwise motions are counted negatively. Theorem 45 W (γ, z 0 ) is an integer If γ and τ are homotopic through closed curves which all miss z 0, then W (γ, z 0 ) = W (τ, z 0 ) The winding number of the circular path γ(t) = z 0 + Re 2πint is n. γ 62

64 CHAPTER 8. WINDING NUMBERS, HOMOLOGY, AND CAUCHY S THEOREM 63 Proof: The second result follows from Cauchy s theorem, and the third is an easy calculation. We need only prove the first item. Let ϕ be a primitive for the curve γ and function z z 0. Then ϕ is locally g γ where dg dz = z z 0, and so constant. eϕ γ z 0 is locally eg z z 0 γ, where d dz ( e g z z 0 ) = 0. It follows that e ϕ γ z 0 Since γ(0) = γ(), we conclude that e ϕ(0) = e ϕ() and therefore e ϕ() ϕ(0) =. So ϕ() ϕ(0) is an integer multiple of 2πi and consequently 2πi γ z z 0 dz = 2πi (ϕ() ϕ(0)) is an integer. Remark: As a corollary, we can prove the fundamental theorem of algebra. There are many proofs of this theorem, but the proof we give is the proof that can be explained to a layman, and thus in a sense the correct proof. Let P (z) = z n + c z n c n be a polynomial over the complex numbers, n > 0. We claim this polynomial takes the value 0. If this is false, then for any positive R, P (Re 2πit ) is a closed path which avoids 0. Also H(t, u) = P (Rue 2πit ) is a homotopy from this closed path to a constant path, and thus W (P (Re 2πit ), 0) = 0. We claim that for R large enough, P (Re 2πit ) is homotopic to Re 2πint by a homotopy avoiding the origin. Since this last path has winding number n, we have a contradiction. The required homotopy is h(t, u) = ( Re 2πit) n + uc ( Re 2πit ) n ucn To complete the proof, we must prove that this express is never zero for R large enough. If it were zero, then ( Re 2πit ) n = [uc ( Re 2πit ) n ucn ] is for some t. The absolute value of the left side is R n and the absolute value of the right side is at most c R n c n max{ c,..., c n }R n, so we have a contradiction as soon as R > max{ c,..., c n }. 8.2 The Inside of a Curve A closed curve γ : [0, ] U is said to be simple if γ does not cross itself, that is, γ(t ) = γ(t 2 ) only if t = t 2 or one of t, t 2 is 0 and the other is. According to the Jordan curve theorem, the complement of a simple curve in the plane has exactly two components, one bounded and one unbounded. The bounded component is called the interior of the curve.

65 CHAPTER 8. WINDING NUMBERS, HOMOLOGY, AND CAUCHY S THEOREM 64 It used to be believed that this theorem was necessary for a rigorous treatment of complex analysis. But we can use the winding number to avoid it. Suppose a curve is continuous and closed, but not necessarily simple. We say that a point z 0 is inside the curve if it is not on the curve and if W (γ, z 0 ) 0. Theorem 46 The complement of a closed curve is a union of open connected components. Moreover. W (γ, z) is constant on components of the complement 2. Exactly one component is unbounded, and W (γ, z) is zero on this component. 3. The union of a closed curve and all points inside it is compact. Remark: For example, consider the curve below. For each component of the complex, compute the Winding Number of the curve about points in that component. Hint: It suffices to fix a point in the component and then create a homotopy of the curve, which doesn t pass through that point, to a much easier curve. Far away curlicues no longer matter. Figure 8.: A Complicated Curve Proof: Connected open sets are arcwise connected. So if z and z 2, are points in the same component, we can find a path τ in this component joining z to z 2. We claim that W (γ, τ(u)) is constant in u, proving the first assertion. This follows from the following principle. The map z z + a translates the entire plane by a. Equivalently, we can leave the plane fixed and translate functions using the formula f(z) f(z a). Then integrating f(z) around γ gives the same result as integrating f(z a) around γ + a, because these equations translate both the function and the path by a. Sincee W (γ, τ(u)) = dζ 2πi γ ζ τ(u), this winding number is equal to W (γ + τ(u), 0). We

66 CHAPTER 8. WINDING NUMBERS, HOMOLOGY, AND CAUCHY S THEOREM 65 now have a homotopy of curves, and so W doesn t change by Cauchy s theorem. Since γ([0, ]) is compact, it is bounded. Hence there is a positive real number R such that {z z > R} does not intersect the curve, and thus must be included in one component of the compliment. That component is the only unbounded component. If z is any point in this infinite ring, W (γ, z) = 0 because γ is homotopic to a constant curve by a homotopy that avoids z, since {z z R} is contractible. The union of the curve and all points inside it is compact because it is closed and bounded. it is closed because its compliment is the union of the remaining open components of C γ. It is bounded because the compliment contains the unbounded component of C γ. 8.3 The Homological Cauchy Theorem Theorem 47 (Homological Cauchy Theorem) Suppose f(z) is holomorphic on a domain U. Let γ,..., γ k be a finite number of closed curves in U. Suppose W (γ i, z 0 ) 0 implies z 0 U. Then γ i f(z) dz = 0. Corollary 48 Let f(z) be holomorphic on a domain U. Let γ be a closed curve in U such that every point inside γ is in U. Then γ f(z) dz = 0. Important Remark: From now on, when we apply this result we will say γ is a closed curve instead of γ,..., γ i are closed curves, and we will say every point inside γ is in U instead of W (γ i, z 0 ) 0 implies z 0 U. Example: The picture below shows a domain U with two holes, and a continuous curve. This curve is not homotopic to a constant in U. However, W (γ, z 0 ) = 0 for points in the holes, because γ goes around each hole once clockwise and once counterclockwise. Consequently, γ f(z) dz = 0 if f is holomorphic on U. This example shows that the homological version of Cauchy s theorem is more powerful than the homotopy version. Figure 8.2: A Curve with Zero Integral.

67 CHAPTER 8. WINDING NUMBERS, HOMOLOGY, AND CAUCHY S THEOREM 66 Example: The illustration below shows an application of the Homological Cauchy Theorem to a region with three boundary curves. If f is holomorphic on the shaded region, the sum of the integrals around these three curves is zero by the theorem, even if f is not defined on the holes, because the sum of the Winding Numbers of the three boundary curves is zero inside either hole. This sort of application will be crucial when we prove the Residue Theorem in a future chapter. Figure 8.3: A Region with Three Boundary Curves Remark: We now prove the theorem. This proof was given around 960 in a complex variable course at Harvard by George Mackey. The details of this proof may be due to Mackey. Proof: By compactness, we can find ɛ such that whenever γ i (t) is a point on one of the curves, the open disk of radius 5ɛ about this point is in U. (The 5 gives us room to wiggle in a minute.) The set of open disks of radius ɛ forms a cover of the union of the curves, so we can find a common decomposition 0 = t 0 < t <... < t n = such that each γ i ([t j, t j ]) is in one such disk of radius ɛ. For each i and j, draw this disk and find a homotopy within the larger disk of radius 5ɛ to a curve with one horizontal and one vertical segment, as illustrated below. Replace the original γ i by the resulting polygonal replacements. Figure 8.4: Small Path Adjustment

68 CHAPTER 8. WINDING NUMBERS, HOMOLOGY, AND CAUCHY S THEOREM 67 We claim it will be enough to prove the theorem for the new curves. Notice that they still satisfy the key assumption that W (γ i, z 0 ) 0 implies z 0 U, for if such a z 0 U, then the homotopy between the original γ i and the new polygonal γ i would avoid z 0 and thus not change the sum of the winding numbers. If we could prove the conclusion of the theorem for the polygonal paths, then any holomorphic function f on U would satisfy γ i f(z) dz = 0 for the new γ i. But again, the homotopies were done within U, so this sum of integrals remains unchanged it we returned to the original γ i. Think of the endpoints of the polygonal paths as vertices of an irregular lattice, and draw a rectangular piece of that lattice. In some cases, horizontal or vertical segments may need to be subdivided to allow this. We will draw the lattice as almost square, but in practice it may be very irregular, with for instance, long narrow rectangles. Below is a picture with two polygonal paths indicated. Note that some of the subrectangles of the diagram will be in U and some not, since U is not shown. Figure 8.5: Main Diagram of the Proof In the above special case, the remaining argument is easy. If z 0 is inside the top right rectangle, then W (γ i, z 0 ) = 0, and so z 0 U. Thus we can apply Cauchy s theorem to conclude that f(z) dz = 0 around this rectangle. The lower left of the diagram shows a path surrounding four rectangles. The winding number sum is for one rectangle, and for the remaining three, so the interiors are in U. Break these paths into two paths. By Cauchy s theorem, both integrals vanish. The

69 CHAPTER 8. WINDING NUMBERS, HOMOLOGY, AND CAUCHY S THEOREM 68 picture suggests more, namely that we could break the bottom left path into a sum of rectangles. That s shown in the full proof. The general case of the proof is really a clever combinatorial argument rather than analysis. Let h ij be formal symbols indicating the horizontal segments with arrows pointing right, and let v ij be formal symbols indicating the vertical segments with arrows pointing up. Form the abelian group G over the integers generated by these symbols. An element of this group is a formal sum m ij h ij + n ij v ij. We can think of elements of this group as general paths, not necessarily connected or closed. Each of our closed polygonal paths σ i induces elements of this group in an obvious way. Notice that the σ i could use individual segments more than once, so elements of the form 3h 2v 23 are possible. We define integration over an element of the group in the obvious way. Each subrectangle of the grid induces an obvious element of G formed by the formal symbols for the four sides of the rectangle with appropriate signs. Denote these elements R ij. Consider the expression W ij R ij G where W ij = W (σ i, z 0 ) for some z 0 inside R ij. In other words, we form a formal sum of rectangles, each occurring as many times as the original polygonal paths went around the rectangle. Notice that only rectangles in U play a role in this expression, by the basic assumption of our theorem. The combinational lemma we need states that this sum of rectangles equals the sum of the original polygonal σ i in G. This lemma will clearly prove the theorem. To prove the result, form σ = W ij R ij formal sum corresponding to polygonal path This is an element of G and we would like to prove that it is zero, i.e., that the coefficients of each h ij and each v ij are zero. We ll give the argument for a horizontal segment h ij. Find a sequence z n converging to the center of this segment from above and a second sequence w n converging to the center from below. Figure 8.6: A Limit

70 CHAPTER 8. WINDING NUMBERS, HOMOLOGY, AND CAUCHY S THEOREM 69 By definition of σ, the integral of z z n over σ and the integral of z w n over σ are both zero. So the integral of z z n z w n over σ is zero. On the other hand, this integrand goes uniformly to zero on all segments except h ij. So if the coefficient of h ij in σ is not zero, we conclude that the limit of is zero. h ij ( z z n z w n ) dz On the other hand, consider the the same integrand over the path below. Figure 8.7: Another Limit By a winding number argument, the value of the integral over this path is 2πi. Again, the integrand converges uniformly to zero on all sides of the rectangle except h ij. So we conclude that the limit of ( ) dz z z n z w n is 2πi. This contradiction proves the result. h ij

71 Chapter 9 Isolated Singularities, Laurent Expansions, and the Residue Theorem 9. The Identity Theorem We mentioned the following theorem in chapter two. It is time to give the details. Theorem 49 (The Identity Theorem) Suppose f and g are holomorphic on a domain U. If U intersects R and f = g on this intersection, then f = g always. More generally, if there is a sequence z n U with a limit z 0 U and f(z n ) = g(z n ), then f = g always Proof: It suffices to prove the second result. Let h(z) = f(z) g(z) and note that h(z n ) = 0. Expand h in a power series about z 0 h(z) = c 0 + c (z z 0 ) + c 2 (z z 0 ) By continuity, h(z n ) h(z 0 ), so h(z 0 ) = c 0 = 0. Let k(z) = h(z)/(z z 0 ) and notice that k(z n ) = 0 and k is holomorphic near z 0 because k(z) = c + c 2 (z z 0 ) +... By continuity, k(z n ) k(z 0 ), so c = 0. Continue. It follows that the power series is identically zero and thus that h(z) is zero on an open neighborhood of z 0. Consider the set V of all points z 0 U such that h = 0 in an open disk about z 0. This set is trivially open; by the previous paragraph it is not empty. We will prove it closed, 70

72 CHAPTER 9. ISOLATED SINGULARITIES AND THE RESIDUE THEOREM 7 so connectedness of U implies that it is all of U. Suppose z n V and z n z 0 U. Then h(z n ) = 0 and the previous argument shows that h is identically zero near z Liouville s Theorem The following beautiful theorem has many consequences. We prove it here because we need it later in the chapter. Definition 3 An entire function is a function defined and holomorphic on the entire complex plane. Theorem 50 (Liouville) A bounded entire function is constant. Proof: Turn back to the proof that a holomorphic function can be expanded in a power series. According to that proof, our entire function has a power series expansion [ ] f(ζ) f(z) = dζ z k 2πi ζk+ k=0 valid in the entire plane. Since power series are given by f (k) (0) z k k! we conclude that k=0 f (k) (0) = k! f(ζ) dζ 2πi γ ζk+ Recall that the absolute value of any complex integral γ f(z) dz is less than or equal to the maximum of f on γ times the length of γ. If f is bounded by B, then the maximum value of f on the circle of radius R about the origin is B. The length of this circle is 2πR. We conclude that f (0) (k) k! B k!b 2πR = 2π Rk+ R k As R, this expression goes to zero for k, so the power series expansion of f about the origin is just f(0). Corollary 5 (The Fundamental Theorem of Algebra) If P (z) is a polynomial of degree at least one, then P has a complex root. Proof: We have already proved this, but Liouville s theorem provides a second proof. If P is never zero, then f(z) = P (z) is an entire function. But ( P (z) = z n + c z n c n z n c z... c ) n z n γ

73 CHAPTER 9. ISOLATED SINGULARITIES AND THE RESIDUE THEOREM 72 and this inequality immediately proves that P goes to zero for large z and thus is a bounded entire function, hence constant, a contradiction. Corollary 52 Let f(z) be a non-constant entire function. Then the range of f is dense in C. Proof: Suppose z 0 is not in the closure of the range. Then constant, which implies that f is constant. f(z) z 0 is bounded, hence Remark: As we remarked earlier, Picard strengthened this by proving that an entire function can omit at most one complex value. We will give his proof later on. 9.3 Isolated Singularities Suppose f and g are holomorphic on a domain U. If g is not identically zero, it is natural to divide and form f g. This function is defined and holomorphic except at zeros of g, and the identity theorem asserts that these are isolated points. So f g is a function with isolated singularities. Definition 4 Let U be an open set containing z 0 and let f be holomorphic on U {z 0 }. Then we say that z 0 is an isolated singularity of f. Remark: There is an amazing theory of isolated singularities. One possibility is illustrated by sin z z. Since z vanishes at the original, the origin is an isolated singularity. But actually, sin z = z z3 3! + z5 5!... = z2 z z 3! + z4 5!... and the singularity can be eliminating by defining the quotient sin z z to be at the origin. We say call such a singularity a removable singularity. Another possibility illustrated by z. Notice that lim z 0 z =. A singularity with this property is called a pole. It turns out that isolated singularities formed as quotients of holomorphic functions like f g are always removable or poles. A third possibility is illustrated by f(z) = e /z. If z n = n, the z n converge to the origin and f(z) 0. On the other hand, if z n = n, the z n 0 and yet f(z n ). So this singularity is not removable and not a pole. It turns out that for any complex number λ, there is a sequence z n 0 with f(z n ) λ. Such a singularity is called an essential singularity. Astonishingly, these are the only possibilities. All sorts of other possibilities suggest themselves. In ordinary calculus, f(x) = x is differentiable except at the origin, but only continuous at the origin. This cannot occur in complex analysis, where isolated points

74 CHAPTER 9. ISOLATED SINGULARITIES AND THE RESIDUE THEOREM 73 of mere continuity are always removable singularities. Similarly, in real variable theory, f(x) = x/ x is differentiable except at the origin, but not even continuous at the origin. This is also impossible in complex analysis; an isolated singularity such that f is bounded nearby is automatically removable. These results will be proved using the theory of Laurent series, discussed next. 9.4 Newton s Calculus Paper In 665, there was a Plague in England and Cambridge University closed for two years. Newton returned to his family s farm, where he had his initial ideas on calculus, gravity, and the theory of light. Newton continued to work on calculus after he returned to Cambridge. He religiously kept his scratch work, which can be found in volume one of Derrick Whiteside s eight volume edition of Newton s Mathematical Papers. In 669, Mercator rediscovered one consequence of Newton s work, and Newton s teacher, Isaac Barrow, told him that he would lose priority if he didn t make his work known. So Newton wrote the first public paper on calculus, De Analysi Per Aequationes Infinitas or On analysis by infinite equations. The paper was not actually published then, but it was deposited in the archives of the Royal Society in London, where visiting scholars including Leibniz read it. Newton preferred to write a complete book on the calculus, which he finished in 673. Unfortunately, complications arose and the book was not published until after Newton s death. De Analysis was published in 7; by that time, Newton was a famous man. The main topic of De Analysi is integration theory, and Newton usually integrates by expanding functions in infinite series and integrating term by term. For example, the first substantial example in the paper is the function y = a2 b+x, which Newton expands by straightforward division. Nowadays we would concentrate on the special case Newton integrates term by term to obtain + x = x + x2 x ln x = x x2 2 + x In private papers, Newton used this result to find ln(.02) to 72 decimal places, and later he wrote to Leibniz I am ashamed of how much time I wasted as a young man calculating functions to many decimal places.

75 CHAPTER 9. ISOLATED SINGULARITIES AND THE RESIDUE THEOREM 74 Immediately after this example, Newton considers y =. This time he notes that +x 2 there are two ways to divide, suggested by writing the denominator as + x 2 or as x 2 +. This gives two series expansions, and when they are integrating term by term, it gives two results + x 2 = x x3 3 + x5 5 x = x + 3x 3 5x 5 + 7x 7... Newton then writes Proceed by the former way when x is small enough, by the latter when it is taken large enough. A few pages later, Newton integrates implicit functions via the same technique. His example is y 3 + axy + x 2 y a 3 2x 3 = 0. Solving for y(x), he obtains y = x 4 a + a2 64x + 3a3 52x a x y(x) = x2 2 a 2 4 ax + 3a3 dx + 64x 52x + 509a x Incidentally, the integral sign is due to Leibniz and Newton didn t know that symbol when he wrote this paper. Newton just says in words the area is, until this last formula. Since he cannot integrate a2 x, he puts a box around the term 64x to indicate that it would have to be evaluated using the numerical techniques from earlier in the paper. There is much more of interest in this first paper. Newton gives ample examples, which makes for easy reading. The surprising thing is that Newton introduces series containing powers of both x and x, which are now called Laurent series, and he already knows that the x term presents special problems, which will lead us to the notion of the residue of a singularity. 9.5 Laurent Series Definition 5 A Laurent Series is an expansion k= c k (z z 0 ) k =... + c 2 (z z 0 ) 2 + c (z z 0 ) + c 0 + c (z z 0 ) + c 2 (z z 0 )

76 CHAPTER 9. ISOLATED SINGULARITIES AND THE RESIDUE THEOREM 75 Remark: Given such a series, we let and f (z) = f 2 (z) = c k z k for z < R k=0 c k z k for z < r k= and call the sum of the series the function ( ) f(z) = f (z z 0 ) + f 2 z z 0 for r < z z 0 < R Thus a Laurent series converges on a (possibly empty) annulus about z 0. For simplicity of notation, we assume z 0 = 0 in the rest of this section, but all of our results obviously apply in the more general case. Example: It is easy to find such series using essentially Newton s method. Take, for instance, the function f(z) = (z )(z 2). This can be rewritten using partial fractions as z + z 2. We can divide the plane into three annuli about the origin: z <, < z < 2, and 2 < z. We get three different Laurent expansions as follows. We have z 2 = z 2 = z = + z + z2 + z z < z = z z 2 z 3... z > ( ) 2 ( z z 2 ) = z 2 3 z 2... z < 2 = z + 2 z z z > 2 2 z and the final series are obtained in the obvious way from these. Theorem 53 If f(z) is holomorphic in an annulus about the origin, there is a Laurent series which converges to f(z) at least in this annulus. This series is unique The series can be differentiated term by term in the annulus The series can be integrated term by term in the annulus, modulo the standard problems with integrating z

77 CHAPTER 9. ISOLATED SINGULARITIES AND THE RESIDUE THEOREM 76 Remark: The proof of existence is similar to the proof that holomorphic functions have power series expansions. Fix z 0 in the annulus. Let g(z) be defined on the annulus by g(z) = f(z) f(z 0) z z 0. This function is holomorphic on the annulus except possibly at z 0. But expanding f as a power series about z 0 gives f(z) = k 0 c k(z z 0 ) k so that k 0 g(z) = c k(z z 0 ) k c 0 = c k (z z 0 ) k z z 0 k Thus g is also holomorphic at z 0. Draw two curves in the annulus as indicated below, and form the integral f(ζ) f(z) dζ 2πi ζ z Figure 9.: Existence of Laurent Series We assume that z is between the two circular paths. Apply the homological Cauchy theorem, noting that the total winding number is zero on the hole inside the annulus. Consequently this integral is zero and f(z) = f(ζ) 2πi ζ z dζ If ζ is on the large circular path γ, then z ζ <, so we can write ζ z = ζ z ζ = k=0 z k ζ k+

78 CHAPTER 9. ISOLATED SINGULARITIES AND THE RESIDUE THEOREM 77 If ζ is on the small circular path γ 2, then ζ z < and we can write ζ z = z ζ z = k=0 ζ k z k+ We conclude that f(z) = k=0 [ ] f(ζ)ζ k dζ 2πi γ 2 z k+ + [ 2πi k=0 f(ζ) dζ γ ζk+ This gives a Laurent expansion for z between the two circles. But we can replace the inner circle by a smaller homotopic circle, and the outer circle by a larger homotopic circle, to get a formula that is valid for more z. By Cauchy s homotopy theorem, the coefficients do not change. We conclude that the Laurent series converges to f throughout the entire annulus where f is defined. ( Proof, continued: To prove the Laurent series unique, suppose f (z) + f ) ( 2 z and g (z) + g ) 2 z are two different Laurent expansions, using the obvious notation for the positive power and negative power terms. Subtracting, 0 = (f g ) + (f 2 g 2 ) and so ( ) (f g )(z) = (f 2 g 2 ) z In this formula, f g is a power series which converges everywhere in the annulus r < z < R and therefore in the disk z < R. Similarly f 2 g 2 is a series in z which converges in the annulus and thus in r < z. Moreover, f 2 g 2 has no constant term and thus goes to zero at infinity. Consequently, the formulas defines an entire function, given by f g on z < R and by (f 2 g 2 ) ( ) z on r < z. This entire function is bounded, and so constant, and so zero. Since a power series expansion is unique, we conclude that the coefficients of the power series for both differences are zero. Proof, part three: Term by term differentiation and integration of Laurent series are proved the same way. We ll discuss the case of differentiation. If f(z) = f (z) + f 2 ( z ) is a Laurent series, the derivative of f equals df dz (z) + df 2 dz ( ) ( z ) z 2 Both f and f 2 are power series which can be differentiated term-by-term. By the displayed formula, the term c k z k in f becomes kc k z k and the term c k z k ( ) z, i.e., c k z k in f 2 becomes kc k z k ( ) ( ) z z, i.e., 2 ( k)c k z k. ] z k

79 CHAPTER 9. ISOLATED SINGULARITIES AND THE RESIDUE THEOREM Classification of Isolated Singularities Laurent series occur most often about an isolated singularity z 0, converging in a punctured disk 0 < z z 0 < R. While the case of more general annuli is sometimes useful, readers should think of the punctured disk about a singularity when they hear the term Laurent expansion. Theorem 54 Let f and g be holomorphic on a domain U and suppose g is not identically zero. Then all isolated singularities of f(z) g(z) are removable or poles. Proof: If z 0 is a removable singularity of the quotient, expand f and g in power series about z 0 : f(z) = (z z 0 ) m f (z) where f and g do not vanish at z 0. Then g(z) = (z z 0 ) n g (z) f g = (z z 0) m n f (z) g (z) If m n 0, the singularity is removable. Otherwise, it is clearly a pole. Theorem 55 Let f(z) be holomorphic on U except for an isolated singularity at z 0. Find the Laurent expansion of f in the annulus 0 < z z 0 < ɛ. Then If the Laurent expansion has only non-negative terms, the singularity is removable. If the Laurent expansion has a finite number of negative terms, including at least one such term, the singularity is a pole. If the Laurent expansion has infinitely many negative terms, the singularity is an essential singularity. Remark: This exhausts the possibilities for a Laurent expansion, and thus the possibilities for an isolated singularity. In particular, if f is bounded near an isolated singularity, then the singularity is automatically removable. Proof: The first result is obvious, since the Laurent expansion is a power series which determines an appropriate value at z = z 0. The second result is also easy. Since the positive terms of the Laurent expansion have a definite value at z 0, we need only consider the negative terms. We want to prove that a non-trivial finite polynomial in z z 0 converges to infinity when z z 0, and this will follow if a non-zero polynomial in z converges to infinity when z.

80 CHAPTER 9. ISOLATED SINGULARITIES AND THE RESIDUE THEOREM 79 Write ( P (z) = z k + c z k c k z k c z c 2 z 2... c ) k z k The required result follows immediately. Finally consider the case of a Laurent series with infinitely many negative terms. Since the positive terms give a definite value at z 0, it suffices to prove that the negative terms can take any limiting value as z n z 0. To simplify notation, we assume z 0 = 0. Suppose the result is false, and the limiting value cannot be λ. Call the sum of the series with negative terms g(z) and let this series converge in 0 < z < B ; choose B < B. Consider g on 0 < z B. If g = λ infinitely often, a subsequence of points where it happens converges and this limit point must be the origin. By assumption, this cannot happen. By shrinking B, we can assume that g is never equal to λ in 0 < z B. Consider h(z) = g(z) λ on 0 < z B. If this function is not bounded near the origin, then there must be a sequence z n with z n < n and > n. So g(z n ) λ < n and again, by assumption this cannot happen. g(z n) λ Therefore, h(z) = g(z) λ is holomorphic in the punctured disk 0 < z < B and bounded near the origin. Expand this function in a Laurent series. By the proof that this can be done, the negative powers of z in this expansion are [ ] h(ζ)ζ k dζ 2πi γ 2 z k+ k=0 Since h is bounded near the origin, and since this formula holds for γ 2 of arbitrarily small radius, we conclude that all of the negative terms vanish, and consequently h can be extended to be holomorphic at the origin. Solving for g(z), g(z) = h(z) + λ. This is a quotient of holomorphic function and therefore had a removable singularity or pole at the origin, contradicting the assertion that it is given by a Laurent series with infinitely many negative powers. Remark: Earlier we proved that the range of a non-constant entire function is dense in C. We now know enough to strengthen this result: Theorem 56 Let f(z) be a non-constant entire function.. If f is a polynomial, then it takes each complex value. But if z n, then f(z n ) 2. If f is not a polynomial, the values of f are dense in the plane. Indeed, for any fixed λ, there is a sequence z n such that f(z n ) λ Proof: If f is not a polynomial, then f ( z ) has an essential singularity at the origin; the second item is a consequence.

81 CHAPTER 9. ISOLATED SINGULARITIES AND THE RESIDUE THEOREM The Residue at an Isolated Singularity Let f(z) be holomorphic on a domain U with an isolated singularity at z 0. The coefficient of z z 0 in the Laurent expansion of f about z 0 is called the residue of f at z 0, and denoted Res(f, z 0 ) Example The residue of cos z z at the origin is because cos z z = z2 2! + z4 4! z = z z 2! + z3 4!... Example 2 Suppose g and h are holomorphic near z 0 and f(z) = g(z h(z). Also suppose that h(z 0 ) = h (z 0 ) =... = h (k ) (z 0 ) = 0, but h (k) (z 0 ) 0. Then Res(f, z 0 ) = k g(k ) (z 0 ) h (k) (z 0 ) The reader should check that this result is true. Example 3 Let f(z) = +z 4. This function has isolated singularities at ±±i 2. The residue at z 0 is 4z 3 0 = 4z 3 0. So the residue at +i 2 = +i 4 2 and the residue at +i 2 = i The Residue Theorem Remark: The following result is the grand climax of our study of Cauchy s theorem. We state it as usually used, and then when multiple closed paths are involved.

82 CHAPTER 9. ISOLATED SINGULARITIES AND THE RESIDUE THEOREM 8 Theorem 57 (The Residue Theorem) Let f(z) be holomorphic on U except at isolated singularities. Let γ be a continuous closed curve in U such that every point inside γ is in U. Assume that no isolated singularity is on γ. Then f(z) dz = (2πi) Res(f, z i )W (γ, z 0 ) γ singularities Figure 9.2: The Residue Theorem Theorem 58 (The Residue Theorem) Let f(z) be holomorphic on U except at isolated singularities. Let γ, γ 2,..., γ k be closed continuous curves in U such that every point inside the γ i is in U, i.e., whenever W (γ i, z 0 ) 0 then z 0 U. Assume no isolated singularity is on a γ i. Then f(z) dz = (2πi) Res(f, z 0 )W (γ i, z 0 ) γi singularities,γ i

83 CHAPTER 9. ISOLATED SINGULARITIES AND THE RESIDUE THEOREM 82 Remark: Note that Cauchy s integral formula is a special case of this theorem. Remark: Here s a small taste of an application. This type of application will be discussed systematically later. Consider the path γ illustrated below. By the residue theorem, dz + z 2 = Res(f, i) 2πi Since γ + z 2 = ( ) = ( ) + higher terms z i z + i z i 2i we have Res i (f, i) = 2i, so the complex integral is (2πi) 2i = π Figure 9.3: Contour Integral One This complex integral is R R dx + x 2 + dz semicircle + z 2 = On the semicircle, z 2 = + z 2 + z 2, so + z 2 z 2 = R 2. So +z. Thus the integral over the semicircle is at most this number times the 2 R 2 πr length of the semicircle, or. This expression goes to zero as R goes to infinity. We R 2 conclude that dx + x 2 = π In this special case, we can directly verify the result because dx + x 2 = arctan x = π

84 CHAPTER 9. ISOLATED SINGULARITIES AND THE RESIDUE THEOREM 83 dx 2 and +i Remark: Here s a second example. We compute of the integrand inside the semicircle are at +i sum of the residues at these points. the same way. The singularities +x 4 2. So the integral is 2πi times the Figure 9.4: Contour Integral Two By a calculation in the previous section, these residues are +i 4 2 ( ) i + i dx + x 4 = (2πi) 4 2 ( i = (2πi) 2 2 and i 4. So 2 ) = π 2 Proof of the Residue Theorem: We ll prove the theorem for one γ; the general proof is exactly the same. As in the diagram below, add paths τ,..., τ k around the isolated singularities z i inside γ, going around W (γ, z i ) times. There are only finitely many isolated singularities inside γ since γ together with its inside is compact. Figure 9.5: Proof of the Residue Theorem

85 CHAPTER 9. ISOLATED SINGULARITIES AND THE RESIDUE THEOREM 84 Apply the homological Cauchy theorem. Since γ goes around z i by W (γ, z i ) times and τ i goes around z i by W (γ, z i ) times and τ j doesn t go around z i at all, the z i are not inside the path formed by γ and the τ i. Thus f is holomorphic inside this path, and Cauchy s theorem gives f(z) dz + f(z) dz = 0 τ j Therefore γ γ f(z) dz = τ j f(z) dz We can shrink the τ j as much as we want without changing the integrals. Eventually τ j will be inside the annulus where the Laurent series for the isolated singularity z i converges. Separate this Laurent series into two parts: all the terms except the z z i term and separately this term, which equals Res(f, z i ) z z i Since the sum of terms of order not can be integrated term by term, the Laurent series has the form dh dz + Res(f, z i) z z i near z i, and dh f(z) dz = τ j τ j dz dz Res(f, z i) τ j dz z z i But dh τ j dz dz = h(end of τ) h(start of τ) = 0. Also dz Res(f, z i ) = (2πi)Res(f, z i )W (τ i, z i ) = (2πi)Res(f, z i )W (γ, z i ) z z i τ j by our choice of τ. In short, f(z) dz = (2πi) Res(f, z i )W (γ, z i ) γ

86 Chapter 0 Fundamental Results We now harvest an array of important results which follow from the Residue theorem and other forms of Cauchy s theorem. 0. Meromorphic Functions Definition 6 A meromorphic function on a domain U is a holomorphic function with isolated singularities, all removable or poles. Remark: We always imagine that all removable singularities are removed, so a meromorphic function has values everywhere. It even has values at poles, if we allow the value to be there. Note that it would make no sense to assign the value to an essential singularity, since nearby values can be anything in that case. Remark: The meromorphic functions on a domain form a field we will sometimes denote M(U). The key point is that f has only zeros and poles if f is not identically zero. It has zeros at the poles of f and poles at the zeros of f. All field axioms are easily checked. Remark: We will focus more and more on meromorphic functions as we proceed; they are the natural objects of study in complex analysis. Notice that from the beginning of these notes on page 8, our examples of holomorphic functions had poles. 0.2 The Order of a Zero or Pole Definition 7 Let f be holomophic on an open neighborhood of z 0, except possibly at z 0. Assume f is not identically zero and z 0 is not an essential singularity. Expand f in a 85

87 CHAPTER 0. FUNDAMENTAL RESULTS 86 Laurent series near z 0, and let k be the lowest power of (z z 0 ) k in this expansion. Then the order of f at z 0 is Ord z0 (f) = k Remark: If f(z 0 ) = 0, then the order of f at z 0 is the multiplicity of this zero, i.e., the number of zeros at z 0. If z 0 is a pole, the order is negative and counts the multiplicity of infinite values at z 0. If z 0 is neither a zero nor a pole, the order of f at z 0 is zero. Remark: Thus order extends the concept of the multiplicity of a zero of a polynomial introduced in algebra. Example: Ord 0 (sin z z) = 3 because sin z z = z3 3! + z5 5!... Theorem 59 The order function has the following properties: Ord z0 (fg) = Ord z0 (f) + Ord z0 (g) Ord z0 (f + g) min {Ord z0 (f), Ord z0 (g)} Ord z0 (f) takes all possible values as f varies. Ord z0 (λ) = 0 if λ is a nonzero constant Remark: The proof is easy. 0.3 The Riemann Sphere It is possible to extend complex analysis to arbitrary surfaces embedded in higher dimensions. When given a complex structure, these objects are called Riemann surfaces. They play a big role in the theory, first implicitly and later explicitly. The simplest of these surfaces is the sphere. Imagine this sphere created by adding a point at infinity to the ordinary plane. Stereographic projection sends the sphere minus its north point to the plane, and the added point at infinity replaces this north pole. Figure 0.: Stereographic Projection

88 CHAPTER 0. FUNDAMENTAL RESULTS 87 We introduce two coordinates on the sphere, called z and w. The z coordinate is the standard coordinate for points on the plane; the added point at infinity has no z coordinate. Non-zero points in the plane also have w coordinates, given by w = z ; we give the point at infinity the w coordinate zero, and the origin in the plane has no w coordinate. The extended plane with these coordinates is known as the Riemann Sphere. We now speak of functions f on the sphere as given by two formulas: f(z) and g(w) = f ( ) w. We can use either of these formulas when speaking of points on the sphere that are not 0 and not. If we are interested in the behavior of a function near the origin, we use f(z). If we are interested in the behavior of a function near, we use g(w) and remove the singularity at w = 0. For example, f(z) = z z 2 is meromorphic on the sphere. It has the value 2 at the origin. It has a zero at and a pole at 2, and its value at infinity is because lim z f(z) =. Surprisingly, there are few meromorphic functions on the Riemann sphere. Theorem 60 The field M(S 2 ) is exactly the set C(z) of all rational functions P (z) Q(z) where P and Q are arbitrary polynomials. Proof: Suppose f(z) is meromorphic on the sphere. Since the sphere is compact, isolated singularities cannot accumulate and thus there are at most finitely many of them. If there are poles of f at z,..., z n of orders k,..., k n, we can form h(z) = (z z ) k... (z z n ) kn f(z) and get an entire function with no poles in the finite plane. It suffices to prove that h(z) is a polynomial, because then f(z) will be a rational function. Since h is meromorphic on the entire Riemann sphere, it has at worst a pole at infinity, and thus g(z) = h( ( ) z has a Laurent expansion with finitely many negative terms near the origin. It follows that h(z) has a Laurent expansion near infinity with only finitely many positive terms. Since h is holomorphic in the entire plane including the origin, this Laurent expansion converges to h everywhere except possible zero. Since f is holomorphic at the origin, the Laurent expansion must extend to the origin and thus must be a polynomial. Remark: Turn back to the end of the previous section on the order function. There we listed some simple properties of Ord. All of these properties except the last make sense for an arbitrary field. Even the last makes sense for the field C(z) of rational functions. It turns out to be relatively easy to prove that the only functions Ord : C(z) {0} Z

89 CHAPTER 0. FUNDAMENTAL RESULTS 88 satisfying these properties are Ord z0 where z 0 is some point on the Riemann surface, including possibly the point at infinity. This fact is exceedingly important in the advanced theory of Riemann surfaces. If we replace the field C(z) by the field Q of rational numbers, the last statement on the list no longer makes sense. Ignore it. The remaining statements make sense. It is possible to find all such Ord maps Q {0} Z. There is one such map for each prime number. Therefore, by some far fetched analogy, the ordinary prime numbers in the case Q correspond to the points on the Riemann sphere in the case C(z). This analogy between number theory and complex analysis continues to hold for deeper and deeper results, and has motivated very important work in the twentieth century. 0.4 The Argument Principle Theorem 6 (The Argument Principle) Let γ be a closed continuous path, and let f be meromorphic on an open set including all points inside γ. Suppose f has no poles or zeros on γ. Then f (ζ) 2πi γ f(ζ) dζ = W(f γ, 0) = W(γ, z i )Ord(f, z i ) z i where the sum is over zeros and poles of f inside γ. Proof: If f has a zero or pole at z 0 of order k, then near z 0 we can write f(z) = (z z 0 ) k h(z) where h is nonsingular and not zero near z 0. Then f f = k(z z 0) k h + (z z 0 ) k h (z z 0 ) k h = k + h z z 0 h near z 0, so the residue at z 0 of f f is k. The Residue Theorem then proves that the left side of our equation equals the right side. We must show that the left and right sides equal the term in the center, which is W(f γ, 0) = 2πi ζ dζ and this will follow if γ f γ f (ζ) f(ζ) = f γ ζ dζ If our path were piece-wise differentiable, this would follow from the chain rule. Since our path is continuous, we must prove this by constructing primitives. If γ : [0, ] C, a

90 CHAPTER 0. FUNDAMENTAL RESULTS 89 primitive for the left integral is a function ϕ(t) on [0, ] which is locally the pullback via γ of a holomorphic h with dh dz = f f. Locally we can find a branch of the logarithm and form Log(f(z)); the derivative of this function is f f. So ϕ is locally the pullback via γ of Log(f(z)), or equivalently the pullback via f γ of Log. And this gives the integral on the right. QED. Remark: Notice that Log(f(z)) = ln f(z) +i arg(f(z)) and d dz Log(f(z)) = f (z). It follows f that γ f dζ measures the increase of the argument of f(z) as we move along γ, hence the name of the theorem. Theorem 62 (Inverse Function Theorem) Let f be holomorphic on a domain U containing z 0 and suppose df dz (z 0) 0. Then there is an open neighborhood V U of z 0 such that f(v) is open f : V f(v) is a homeomorphism f : f(v) V is holomorphic Proof: Without loss of generality, we can assume that z 0 = 0. Choose a small counterclockwise circle γ about the origin which does not go around any other zero of f. The argument principle gives 2πi γ f dζ = W(f γ, 0) = f Since the winding number is constant on open connected components of the compliment of paths, we can find a small disk W about the origin inside f γ such that W (f γ, λ) = for all points in this disk. f(z) Figure 0.2: Proof of Inverse Function Theorem By definition, this winding number equals 2πi f γ ζ λdζ, and as in the proof of the argument principle, this equals 2πi f λdζ. By the Residue theorem, this equals the f number γ

91 CHAPTER 0. FUNDAMENTAL RESULTS 90 of times f = λ inside γ minus the number of poles inside γ. So f = λ exactly once inside γ and f at the point where this happens is not zero. Let V = f (W) (interior of γ). Then f : V W is one-to-one and onto. We want to prove that f is a homeomorphism from V to W and that f has a derivative at each point of W. Formally, the argument for a derivative goes like this: f (w) f (w 0 ) f (f(v) = f (f(v 0 )) lim = lim = lim w w 0 w w 0 f (v) f (v 0 ) f(v) f(v 0 ) ( v v0 f(v) f(v0 ) v v 0 ) = Since each w comes from a unique v and f exists and is never zero on V, this formal argument will follow if w w 0 implies v v 0, and this will follow from the continuity of f. So the entire theorem is proved modulo this continuity. But the continuity of f was essentially proved by our earlier argument. Indeed, we know that f is not zero at each point v of V. The previous argument then showed that we can choose an arbitrarily small open disk about v such that the image of this disk contains an open neighborhood of f(v), and we can then shrink the arbitrarily small open disk to an even smaller open set about v whose image is this open neighborhood of f(v). Since any open set X in the domain can be covered by open disks inside X, and thus by the even smaller open sets with open image, the image of X in W must be open. f (v 0 ) 0.5 The Open Mapping Theorem The inverse function theorem together with our study of conformal maps in chapter two gives a complete picture of holomorphic maps near z 0 with f(z 0 ) 0. Namely, we can find an open neighborhood V of z 0 and an open neighborhood W of f(z 0 ) and then f : V W and f : W V are differentiable homeomorphisms which preserve angle. We are going to extend this picture to points z 0 where f (z 0 ) = 0. Suppose the order of the zero of f at z 0 is k, where k and thus k 2. Then near z 0 we have f(z) = f(z 0 )+ f (k) (z 0 ) (z z 0 ) k +... = f(z 0 )+(z z 0 ) k ( c k + c k+ (z z 0 ) + c k+2 (z z 0 ) ) k! Note that h(z 0 ) 0. So we can define a branch of the logarithm Log near h(z 0 ) and write h(z) = ( e Log(h(z))/k) k = g(z) k where g is holomorphic near z 0 and g(z 0 ) 0. Therefore f(z) = f(z 0 ) + ((z z 0 )g(z)) k. Notice that (z z 0 )g(z) is defined near z 0 and has non-zero derivative at z 0. By the inverse function theorem, this is a homeomorphism near z 0 with holomorphic inverse. By shrinking the image set, we can suppose that it is a disk about the origin.

92 CHAPTER 0. FUNDAMENTAL RESULTS 9 We have proved Theorem 63 Suppose f is holomorphic on U and f has a zero at z 0 of order k > 0. Then f is given by the following composition of maps near z 0. The first map is a translation, the second is a homeomorphism, the third is z z k and the final map is another translation. So up to homeomorphisms, f locally wraps a neighborhood around z 0 k times. Figure 0.3: Conformal Maps Near a Zero of df dz Corollary 64 (The Open Mapping Theorem) Suppose f is a non-constant holomorphic map on a domain U. Then f is an open mapping; it maps open sets to open sets. Proof: It suffices to prove this locally, but points where f (z 0 ) 0 have arbitrarily small open neighborhoods which map to open sets by the inverse function theorem, and this remains true if f (z 0 ) = 0 by the above sequence of maps. Corollary 65 A non-constant holomorphic function on a domain U cannot be real-valued, or take values only on an arbitrary line in the plane, or take values only on the unit circle. 0.6 The Maximum Principle Theorem 66 (The Maximum Principle) The absolute value of a non-constant holomorphic function on a domain U cannot have a local maximum.

93 CHAPTER 0. FUNDAMENTAL RESULTS 92 Proof: This is a trivial consequence of the open mapping theorem. Corollary 67 Let f and g be continuous on a closed disk and holomorphic on the interior of this disk. If f = g on the boundary of the disk, then f = g everywhere inside. Proof: Consider h = f g. Suppose h = 0 on the boundary of the disk. Then either f = g everywhere or else f(z) g(z) has a maximum value inside the disk, which cannot happen by the previous result. 0.7 Uniform Limits of Holomorphic Functions The following amazing theorem is false for real-valued functions, as illustrated by the pictures below. Figure 0.4: Uniform Limits in Real Variables Theorem 68 Suppose f n are holomorphic on a domain U and f n (z) f(z) uniformly on each compact subset of U. Then f(z) is holomorphic and dfn dz df dz uniformly on each compact subset of U. Consequently all higher derivatives of f n also converge to the corresponding higher derivative of f uniformly on compact subsets. Proof: To prove that f is holomorphic, it suffices by Morera s theorem to prove that R f(z) dz = 0 for all rectangles contained, together with their interiors, in U. But this holds for f n since f n is holomorphic. Moreover, by uniform convergence, R f n(z) dz f(z) dz. R Next we prove that uniform convergence of functions implies uniform convergence of derivatives. This follows from the following formula for the derivative of f, a generalization of Cauchy s formula which follows from the Residue theorem: if γ is a small circle about

94 CHAPTER 0. FUNDAMENTAL RESULTS 93 z, df dz (z) = f(ζ) 2πi γ (ζ z) 2 dζ Suppose this circle has radius R. Then for z on or inside the disk of radius R/2, we have df n dz df dz max γ f n (z) f(z) R/2 (2πR) Since f n f uniformly on the circle of radius R, df n dz df dz uniformly on the interior of a disk of radius R/2. finitely many such disks, so we are done. Any compact set can be covered by

95 Chapter Automorphism Groups of the Complex Plane, the Unit Disk, and the Riemann Sphere This chapter is the first of three which lead to the Riemann Mapping Theorem, the central result of Riemann s PhD thesis. Results in the preliminary chapters are of independent interest. If U is a domain in C or the Riemann Sphere, an automorphism of U is a one-to-one and onto holomorphic map ϕ : U U. It is easy to see that the automorphisms of U form a group. We will compute these groups for the open unit disk D, the entire complex plane C, and the Riemann sphere S 2. It turns out that these are the only simply connected Riemann surfaces, although we will not prove that. So our examples were not chosen at random.. The Automorphism Group of C Theorem 69 The automorphisms of C are precisely the maps where a and b are complex and a 0. ϕ(z) = az + b Proof: Since these are clearly automorphisms, it suffices to prove that all automorphisms have this form. An automorphism is an entire function, so by theorem 54, it is either a 94

96 CHAPTER. AUTOMORPHISM GROUPS 95 polynomial or else for each fixed λ, we can find z n with ϕ(z n ) λ. But ϕ is onto, so there exists a z 0 with ϕ(z 0 ) = λ. Since ϕ is one-to-one, the derivative of ϕ at z 0 cannot be zero, so the inverse function theorem states that there bounded open neighborhoods W and W 2 of z 0 and λ such that W is mapped one-to-one and onto to W 2 by ϕ. The values in W 2 are only taken by ϕ in W, so λ cannot be a limiting value of ϕ near infinity. It follows that ϕ is a polynomial. Since ϕ is one-to-one, this polynomial cannot have multiple roots or one root of order greater than one. So it is linear. QED. Remark: The complex plane has a natural metric. Isometries, that is, distance preserving maps, must be conformal by the law of cosines. So our group contains all isometries of the plane. These have the form ϕ(z) = e iθ z + b If e iθ =, these are translations. Otherwise they are rotations about some point. Indeed, any map ϕ(z) = az + b with a has a fixed point because az + b = z is solvable, and it is easy to see that when a = e iθ, the resulting map is rotation about this fixed point. The remaining maps have the form ϕ(z) = me iθ z + b for m > 0; m. Such maps have fixed points; they magnify about the fixed point by m and then rotate by θ. The globally conformal maps which are not isometries are therefore, essentially, magnifications..2 The Automorphism Group of the Riemann Sphere Definition 8 A linear fractional transformation is a map with a, b, c, d complex and ad bc 0. ϕ(z) = az + b cz + d Remark: If we multiply each of a, b, c, d by the same non-zero constant λ, we get the same map. It follows that the set of linear fractional transformations form an object of complex dimension 3 and so real dimension 6. Notice that the condition ad bc 0 does not change the dimension. Theorem 70 The automorphisms of the Riemann Sphere are precisely the following linear fractional transformations, with ad bc 0: ϕ(z) = az + b cz + d

97 CHAPTER. AUTOMORPHISM GROUPS 96 Proof: We leave it to the reader to show that these maps are automorphisms. Notice that z = d c maps to, and maps to a c. Linear fractional transformations are 3-transitive: any three points can be mapped to any other three points by exactly one ϕ. For instance, ϕ(z) = C B C A z A z B maps A to 0, B to, and C to. Now let ϕ be an arbitrary automorphism of the sphere. Multiplying by a suitable linear fraction transformation, we can suppose that 0 0,,, and it suffices to show that this composition is the identity. Since maps to, no other point maps to infinity and our map is an automorphism of C, and hence has the form ϕ(z) = az + b. Since 0 0, b = 0. Since, a =. QED. Theorem 7 A linear fractional transformation is either of the form az + b with a 0 or of the following form with c 0, B = a c, A = ad bc c : (Az + B) z (cz + d) Proof: A simple calculation. Remark: It follows that linear fractional transformations are products of translations of the plane, rotations about a point in the plane, magnifications about a point in the plane, and the map z. This last map is particularly interesting; it is essentially inversion in a circle. Suppose p is a point in the plane and R is a fixed positive number. We define inversion in the circle of radius R about p to be the map which interchanges p and, fixes the circle pointwise, and maps the inside to the outside and the outside to the inside by sending q p to the point Q on the ray from p through q, where q p Q p = R 2. So this map turns the circle inside out. Inversion in a circle reverses orientation, so it is cannot be holomorphic. Actually inversion in a circle of radius about the origin is given by z z because the product of the z 2 z absolute value of z and the absolute value of is one and z is on the same ray through z 2 z 2 the origin as z. Our map z z equals ϕ(z) = z = z zz = z z 2 = z z 2 z and thus is reflection across the x-axis followed by inversion in the circle. The inclusion of this reflection produces an orientation-preserving map.

98 CHAPTER. AUTOMORPHISM GROUPS 97 Figure.: Inversion in a Circle The above picture shows a pure inversion in a circle without the additional reflection. We invert in the black circle. The red circle inside moves to a bigger red circle outside. The blue circle meets the circle of inversion perpendicularly, and a little thought shows that it consequently maps to itself. The green circle goes through the origin, so its image goes through and becomes a straight line. Remark: More generally, if we have a sphere S n with center p and radius R in R n, the map inversion in the sphere interchanges p and, fixes the sphere pointwise, and sends q on the ray from p to q to the point Q on the same ray, where q p Q p = R 2. It is not difficult to show that this map is conformal. With somewhat more difficulty, one can prove Theorem 72 If ϕ : U V is one-to-one, onto, and conformal, where U and V are domains in R n and n 3, then ϕ is a composition of translations, rotations, magnifications, reflections, and inversions in spheres. Notice that the previous result is a local result. In contract, we know that there are enormously many local conformal maps in R 2. So the proper way to generalize complex analysis to higher dimensions is certainly not to concentrate on conformal maps as the central idea of the subject. Theorem 73 A linear fractional transformation maps circles to either circles or lines, and lines to either circles or lines.

99 CHAPTER. AUTOMORPHISM GROUPS 98 Proof: This result is obviously true for translations, rotations, and magnifications, since these maps send circles to circles and lines to lines. By theorem 69, it suffices to prove the result for z = x iy x 2 +y 2. The general equation of a circle or straight line in the plane is The image or equivalently A(x 2 + y 2 ) + Bx + Cy + D = 0 ( ) x y, satisfies this equation provided x 2 +y 2 x 2 +y 2 ( x 2 + y 2 ) A (x 2 + y 2 ) 2 + B x x 2 + y 2 + C y x 2 + y 2 + D = 0 A + Bx Cy + D(x 2 + y 2 ) = 0 Both of these equations have the same form. Indeed, the possible solutions of the first equation are a circle, a line, a point, the empty set, or everything, and similarly for the solutions of the second equations. Since our linear fractional transformation is one-to-one and onto, one equation has solution a circle or line just in case the other equation has tone of these solutions. Remark: A calculation shows that stereographic projection, illustrated on page 8, is a conformal map from the sphere to the plane. Give the sphere the standard Riemannian metric; then isometries of this sphere are conformal and map to linear fractional transformations. The site rotation.pdf contains a paper by Joel H. Shapiro proving that the linear fractional transformations corresponding to isometries of the sphere are exactly those induced by a unitary matrix ( a b ) c d Note that the matrix induced by az+b cz+d is only unique up to multiplying all components by the same nonzero λ. The paper s assertion means that a linear fractional transformation induces an isometry of the sphere if and only if one representative matrix for the transformation is unitary. Notice that z comes from an isometry of the sphere. This is a coincidence, and inversions in other circles do not. Example: It is traditional to stop here and draw many wonderful pictures of maps obtained from linear fractional transformations. We ll briefly consider one case. Let ϕ(z) = i z + z

100 CHAPTER. AUTOMORPHISM GROUPS 99 This maps sends, 0, and i. Thus the maps sends the circle through and and i to the line through and 0 and, i.e., the x-axis. It follows that this map is a conformal isomorphism from the interior of the unit disk to the upper or lower half plane bounded by the x-axis. Since 0 i, it maps the interior of the unit circle to the upper half plane. Notice that the inverse of this map is ϕ (z) = + iz iz.3 The Automorphism Group of the Open Unit Disk We begin with a beautiful, and very important, lemma Lemma 6 (Schwarz Lemma) Let f(z) be a holomorphic map from the open unit disk to itself and suppose that f(0) = 0. For each z in the disk, f(z) z If this inequality is an equality for some nonzero z, then f(z) is a rotation, f(z) = e iθ z. Proof: Apply the principle of the maximum to g(z) = f(z) z. On the circle of radius ɛ, g is at most one and z = ɛ, so g ɛ. By the maximum principle, g is at most this value everywhere inside this circle of radius ɛ. As the circle grows closer to, the bound grows tighter, so ultimately we find that g inside the disk. If g = at some z inside the disk, then g takes its maximum value inside the disk and thus is constant by the maximum principle. Corollary 74 Let f be an automorphism of the unit disk mapping the origin to the origin. Then f is rotation, f(z) = e iθ z. Proof: Apply the Schwarz lemma to f and f. We conclude that f(z) z and f(z) z. Hence f(z) = z and the result follows from the Schwarz lemma. Theorem 75 The automorphism group of the open unit disk D is the set of all as z 0 D and 0 θ < 2π. z z 0 f(z) = e iθ z z 0 z z 0 Proof: We will prove that z z 0 maps the unit circle to itself. Since the origin maps to z 0 and z 0 <, this linear fractional transformation must map the interior of the circle

101 CHAPTER. AUTOMORPHISM GROUPS 00 to the interior and thus be an automorphism of the unit disk. It is enough to prove that when z is on the unit circle, z z 0 2 = z z 0 2 or or (z z 0 )(z z 0 ) = ( z z 0 )( z z 0 ) zz z 0 z zz 0 + z 0 z 0 = zz 0 zz 0 + zzz 0 z 0 and this holds because the middle terms cancel and zz = and z 0 z 0 = zzz 0 z 0. If ϕ is an arbitrary automorphism of the unit disk, it maps 0 to some z 0 in the disk. Then z z 0 z z 0 ϕ (z) is an automorphism which maps 0 to 0, so it equals e iθ z for some θ, and thus or equivalently z z 0 z z 0 ϕ (z) ϕ(z) = e iθ z ϕ(z) e iθ z z 0 z z 0 = ϕ(z) Theorem 76 The conformal isomorphism ϕ(z) = i z +z from the unit disk to the upper half plane maps the automorphisms group e iθ z z 0 z z 0 to the automorphism group az+b cz+d of linear fractional transformations with real coefficients a, b, c, d such that ad bc > 0. Remark: This group is usually called SL(2, R)/±I since multiplying a, b, c, d by ± does not change the linear fractional transformation. Serge Lang wrote an entire book on SL(2, R). Figure.2: Disk is Conformally Equivalent to Upper Half Plane

102 CHAPTER. AUTOMORPHISM GROUPS 0 Proof: Note that the conformal map from the unit disk to the upper half plane is linear fractional, and the automorphism group of the unit disk is linear fractional. It follows that the automorphism group of the upper half plane contains only linear fractional transformations. Such a transformation preserves the upper half plane if and only if it maps the real line to itself, and does not interchange upper and lower half planes. Suppose az+b cz+d is such a transformation At least one of a, b, c, d is non-zero. Say it is a and pick λ such that λa is real. Replace a, b, c, d by λa, λb, λc, λd. If z and az+b cz+d are both real, then az + b cz + d = ( ) az + b = a z + b cz + d c z + d = a z + b c z + d Thus the right and left sides of this equation are linear fractional transformations which agree on all real numbers, and thus on three numbers, and hence are equal up to a constant multiple. Since a is real and nonzero, a = a and the constant multiple is one. Hence a, b, c, and d are all real. Conversely, if a, b, c, d are real and ad bc > 0, then az+b cz+d preserves the real line and thus maps the upper half plane conformally to either the upper half of lower half plane. But ai+b ci+d is (ai + b)( ci + d) (ad bc)i ac + bd = (ci + d)( ci + d) c 2 + d 2 + c 2 + d 2 and this is in the upper half plane if and only if ad bc > 0. Remark: Notice that the automorphism groups of the plane, disk, and sphere are all transitive; every point can be mapped to every other point. This is highly unusual and almost never happens for other Riemann surfaces..4 The Unit Disk and Non-Euclidean Geometry The results in this section, while important, play no role in the proof of the Riemann mapping theorem. Remark: The plane and sphere have natural metrics; orientation preserving isometries are conformal. The plane has additional conformal automorphisms (magnifications) and and so does the sphere (inversions in circles). It is a remarkable fact that the disk has a natural metric, and this time all conformal maps are orientation preserving isometries. The metric is exactly the non-euclidean geometry discovered by Lobachevsky, Bolyai, and Gauss, although the connection between geometry and complex analysis was not made until toward the end of the 800 s, by Poincare.

103 CHAPTER. AUTOMORPHISM GROUPS 02 We sketch the resulting theory. According to Riemann, a metric is best given as a quadratic form ds 2 = g ij (x, x 2,..., x n )dx i dy j This expression means that we can find lengths of curves (γ (t),..., γ n (t)) for a t b by computing b gij(γ (t),..., γ n (t)) dγ dt... dγ n dt dt a Curves that are locally the shortest between two points are called geodesics and these geodesics ( replace the ) straight lines of classical geometry. Curves γ have tangent vectors X = dγi dt,..., dγn dt and the inner product between such tangent vectors is given by < X, Y >= g ij X i Y j from which angles of intesection of curves can be computed by the standard formulas. For instance, in R n we set g ij = δ ij and recover the usual formulas of advanced calculus. We now introduce the metric of non-euclidean geometry. The formula is simplest in the upper half plane, where it is ds 2 = dx2 + dy 2 y 2 In the unit disk, this becomes ds 2 dx 2 + dy 2 = 4 ( (x 2 + y 2 )) 2 These formulas say that distance over small regions is essentially computed by the standard formula from Pythagoras; however in the case of the upper half plane, distances very close to the x-axis are much larger than they look, and distances high above the x axis are much smaller than they look. In the case of the disk, the formula says that distances close to the origin are as expected, but distances close to the boundary are much larger than they look. The constant 4 has been added to make the Gaussian curvature equal. The crucial assertion is that our conformal automorphism groups preserve these metrics and become isometries of non-euclidean geometry. Let s prove that. It is easiest to work in the upper half plane where the group elements are az+b cz+d for a, b, c, d real, and ad bc =.

104 CHAPTER. AUTOMORPHISM GROUPS 03 We ll give the essential calculation, letting the reader disentangle that it really works. Let I(z) be the imaginary part of z. Then ds 2 = dx2 + dy 2 y 2 = (dx + idy)(dx idy) I(z) 2 = dz dz I(z) 2 Suppose w = az+b (cz+d)a (az+d)c cz+d. Then dw = dz = (cz+d) 2 (cz+d) 2 Also I(w) = 2 Consequently dw dw = ( az + b cz + d az + b ) = cz + d 2 = ( (ad bc)(z z) 2 cz + d 2 dz dz cz + d 4 dz, so ( ) (az + b)(cz + d) (az + b)(cz + d) ) = cz + d 2 ( I(z) cz + d 2 ( ) ( ) dw dw dz dz cz + d 4 I(w) 2 = cz + d 4 = I(z) ) dz dz I(z) 2 Remark: This result will imply that the conformal automorphisms of the unit disk preserve the Poincare metric on that disk if we can prove that the conformal map from the upper half plane to the unit disk, given by z +iz iz, maps the Poincare metric on the upper half plane to the Poincare metric on the unit disk. The metric on the upper half plane is and the metric on the unit disk is first to the second. As before, and therefore ( dw dw = dw = 2i ( iz) 2 4dz dz ( z 2 ) 2 ( iz)i ( + iz)( i) ( iz) 2 = and we want to prove that w = +iz iz ) ( ) 2i + iz) 2 dz dz = 2i ( iz) 2 dz 4 ( + 2I(z) + z 2 dz dz ) 2 dz dz I(z) 2 carries the Now w 2 = + iz iz 2 = ( + iz iz ) ( ) iz = + iz + iz iz + z 2 2I(z) + z 2 = iz + iz + z 2 + 2I(z) + z 2

105 CHAPTER. AUTOMORPHISM GROUPS 04 and ( w 2 ) 2 = ( ) 2 ( 2I(z) + z 2 + 2I(z) + z 2 = 4I(z) + 2I(z) + z 2 ) 2 Putting these two results together, we have 4 dw dw ( w 2 ) 2 = 6 ( + 2I(z) + z 2 dz dz ) 2 ( ) + 2I(z) + z 2 2 = 4I(z) dz dz I(z) 2 Remark: These calculations show that complex analysis on the unit disk is closely related to Non-Euclidean geometry, because the isometries of that geometry are just the complex automorphisms of the disk, and in particular are all linear fractional transformations. Figure.3: Escher s Angels and Devils Remark: Escher s wonderful work Angels and Devils is a picture of Non-Euclidean geometry. In the picture, each angel and devil has the same non-euclidean size. As we approach the boundary, the angels and devils only seem to grow smaller; note that there are infinitely

106 CHAPTER. AUTOMORPHISM GROUPS 05 many of them. Some people believe that this picture was drawn on a sphere and then projected downward; that is certainly false because a sphere could contain only finitely many angles. Remark: Let us determine the straight lines in this geometry, that is, the geodesics. We first claim that straight lines through the origin are geodesics. This becomes clear when we write the metric in polar coordinates: (dr) ds = (rdθ) 2 r 2 If we want to go from the origin to the boundary, the r variable must increase from r = 0 to r equal to one; any wiggles in θ will only increase the integral for curve length. Figure.4: Poincare s Model for Non Euclidean Geometry Once we know that lines through the origin are geodesics, it follows that images of these lines under automorphisms of the disk are also geodesics. The authomorphism of the disk are all linear fractional, and thus send lines and circles to lines and circles. Since straight

107 CHAPTER. AUTOMORPHISM GROUPS 06 lines through the origin meet the boundary circle perpendicularly, this must still hold after we map by an automorphism. We conclude that Theorem 77 The geodesics of the Poincare model of Non-Euclidean geometry are straight lines through the origin and circles which meet the boundary at 90 degrees. Remark: The previous illustration shows several of these geodesics. They are also visible in Escher s drawing. Notice that lines through the origin alternately bisect angels and devils. Notice that other curves can be seen alternately bisecting angels and devils. They are all circles meeting the boundary at 90 degrees. We gather below other results from Non-Euclidean geometry. Start with parallel lines. Select a line L and a point p L. Drop a perpendicular from p to q L. At p, draw the line M perpendicular to this perpendicular. This new line is a parallel to L through p. See the illustration below. Figure.5: Parallel Postulate In Non-Euclidean geometry, the sum of the angles of a triangle is always less than π, and consequently the sum of the angles of a quadralateral is always less than 2π. From this, it follows that no other line except the line from p to q is perpendicular to both L and M. In the diagram, two red lines are shown. These are called limiting parallels; they are parallel to L and the lines through p between these two are exactly the lines through p parallel to L. In L and M are arbitrary parallel lines in Non-Euclidean geometry, either there is exactly one line segment between them that is perpendicular to both, or else there is no such line segment and the two lines are limiting parallels.

108 CHAPTER. AUTOMORPHISM GROUPS 07 There is a beautiful formula relating the Non-Euclidean length a of the segment pq and the angle of parallelism Π, defined as the angle between pq and the first limiting parallel: cos Π = tanh a = ea e a e a + e a This result was proved in 840 by Lobachevsky. If a is very small, tanh a is almost zero and cos Π is almost zero, so Π is almost π 2. Of course in Euclidian geometry this angle is exactly π 2. On the other hand, as a, tanh approaches and so cos Π and Π is almost zero. Imagine that the universe is Non-Euclidean, but very large, so for distances a we are likely to encounter, Π is almost π 2 and the angle between limiting parallels is so small it would be undetectable. Incidentally, Lobachevsky s formula shows that there is a natural unit of length in Non- Euclidean geometry, since a is measured in this unit and Π is measured in radians, as is usual for angles. There are no magnifications in Non-Euclidean geometry. We already know this another way, because the group of conformal automorphisms contains only isometries in the disk case. If the universe were Non-Euclidean, we could determine this natural unit of length by experiment. It would be very, very, very large. Remark: Next we discuss circles. It is clear that a circle about the origin in the disk is a standard circle. But then a circle about any other origin is a standard circle because the isometry group maps circles to circles or lines. (The center of this circle isn t in the usual position.) The circumference of a circle with Non-Euclidean radius r is ) C = 2π sinh r = 2π (r + r3 3! + r5 5! +... and the area of a circle with Non-Euclidean radius r is ( ) r 2 A = 2π(cosh r ) = 2π 2! + r4 4! +... Remark: Next a few remarks on Non-Euclidean triangles. The sum of the angles of any triangle is smaller than π. But more is true: Theorem 78 The area of a triangle with angles α, β, γ is π (α + β + γ) The extremal case is illustrated be- It follows that all triangles have area less than π. low.

109 CHAPTER. AUTOMORPHISM GROUPS 08 Figure.6: Limiting Triangle with Maximum Area Remark: Similar remarks hold for quadrilaterals, and other polygons. Remark: In Euclidean geometry, three distinct points in the plane determine a unique line or circle. This theorem is still true for the unit disk in the Euclidean plane, of course. But in terms of Non-Euclidean geometry, there is a puzzle: the curve determined by three points could be a line through the origin, or a circle which meets the boundary at 90 degrees, or a circle entirely inside the disk. But it could also be a line which does not go through the origin, or a circle which meets the boundary at two points but not with angle 90 degrees, or a circle which meets the boundary at just one point. The first three of these curves are Non-Euclidean lines and circles. The last three aren t. Suppose L is a non-euclidean line. At each point of this line, erect a perpendicular of length d. Connect the ends of these perpendiculars. This gives a new curve, and in Non- Euclidean geometry it is not a straight line. Such curves are called equidistant curves. In Poincare s model, these curves are straight lines or circles which meet the boundary at two points, but not at 90 degrees. Curves which are circles meeting the boundary at exactly one point are called horocyles; these can be characterized with an internal property. So three points in Non-Euclidean geometry determine a unique curve, which is either a straight line, a circle, an equidistance curve, or a horocycle.

110 Chapter 2 A Topology on the Set of Holomorphic Maps If a power series c n z n has radius of convergence R, then the partial sums f n (z) converge to the total sum f(z) uniformly on compact subsets of the open disk of radius R, since they converge uniformly on closed disks of radius R < R. At the end of chapter ten, we discovered that the topology of uniform convergence on compact subsets behaves unusually well for holomorphic functions. For instance, if f n f in this sense and each f n is holomorphic, then the sum is holomorphic and dfn dz df dz, again uniformly on compact subsets. If U is a domain, let H(U) be the set of all holomorphic functions on U and let C(U) be the set of all continuous complex-valued functions on U. We want to formalize the previous remarks by making H(U) and C(U) into topological spaces with the property that f n f means f n converges to f uniformly on compact subsets of U. Topological spaces induced by metrics have reasonable properties, but more general topologies can behave in unexpected ways. In particular, convergence of sequences determines the topology on metric spaces, but not in general. Therefore, we are going to produce a metric on our function spaces which induces the required meaning of f n f. We will never refer to the metric again; it s sole purpose is to allow us to use topological arguments without worrying about subtle points. 09

111 CHAPTER 2. TOPOLOGY ON H(U) 0 2. A Metric on H(U) and C(U) It is easy to find a countable number of compact subsets K i U whose interiors cover U. For instance, we can select closed disks inside U centered at rational points and with rational radii. If f and g belong to either of our function spaces, define d(f, g) = i min [max Ki ( f g ), ] 2 i This sum converges because 2 i converges. It is symmetrical in f and g. If d(f, g) = 0, then each term must be zero and therefore f = g on each K i ; since the interiors of the K i cover U, f = g everywhere. Finally, the triangle inequality holds because it holds for the numerators of each term. Therefore, d makes H(U) and C(U) into metric spaces. Suppose f n f in this metric, and suppose K is compact. Since the interiors of the K i cover U, we can find a finite number of K i with K (K... K n ). Let ɛ be given. Choose ɛ smaller than min ( 2, ɛ ) n 2. Choose N so that for n > N we have n d(fn, f) < ɛ. This inequality will fail if we use in any of the numerators of the first n terms of the distance formula, so for i n we have max K i ( f n f ) < ɛ 2 i < ɛ 2. Consequently, f n n f < ɛ on (K K 2... K n ) and thus on K. It follows that f n f uniformly on compact subsets of U. Conversely, suppose f n f uniformly on compact subsets of U. Let ɛ > 0 be given. Choose n such that i>n < ɛ 2 i 2. Let K = (K K 2... K n ). Choose N so large that n > N implies f n f < ɛ 2n on each K i. Then n > N implies d(f n, f) < ɛ. Therefore f n f in the metric. Remark: In the next few chapters we will construct important holomorphic functions using the topology of uniform convergence on compact subsets. Our f n (z) will be more complicated than mere powers of z, and the domain U will no longer be a disk. However, we sometimes obtain f(z) in a deeper manner, by using compactness arguments on special subsets of H(U). These arguments depend on a characterization of compactness we give next. This characterization uses deep properties of holomorphic functions and definitely is false for continuous functions or differentiable functions of a real variable.

112 CHAPTER 2. TOPOLOGY ON H(U) 2.2 The Key Technical Theorem Theorem 79 (Montel) Let D be a closed disk contained in an open domain V. Let f n be a sequence of holomorphic functions on V. Suppose that this set is uniformly bounded on D, that is, suppose there is a bound B such that for all n we have f n B on D. Then there is a subsequence of the initial sequence which converges uniformly on compact subsets of the interior of D. Remark: This theorem is definitely false when we replace H with C, the space of continuous functions, or C, the space of C functions. It doesn t matter if the domain of the functions is a subset of R, or of C, or of a more general space. For instance, suppose f n (x) = for negative x, f n (x) is a straight line from (0, ) to (/n, 0) on [0, /n], and then f n (x) = 0 as shown below. Clearly no subsequence converges uniformly. We could round the corners to get the same example for differentiable functions, and round more subtly to get it for C functions. Figure 2.: Montel s Theorem is False for Real Variables Proof: Let γ be the counterclockwise boundary of the disk. If f is holomorphic in a neighborhood of the disk, then the derivatives of f at the origin are given by the generalized Cauchy formula f (k) (0) = k! f(ζ ) dζ 2πi γ ζk+ Suppose the radius of the disk is R. Apply this result to each f n in our sequence to conclude that f n (k) (0) k! B Bk! 2πR = 2π Rk+ R k Since this bound does not depend on n, the values of f n (0) are bounded and we can find a subsequence of the f n such that these values converge to a value we will call f(0). Since the dfn dfn dz (0) are bounded, we can find a subsequence of this subsequence so the dz (0) also converge to a value we will call df dz (0). Continue. By a standard diagonal argument, we

113 CHAPTER 2. TOPOLOGY ON H(U) 2 can find a subsequence of the original sequence which is ultimately a subsequence of each of these subsequences, and therefore such that f (k) n (0) f (k) (0) for all k. We now claim that the series f(z) = k k! f (k) (0)z k = c k z k has radius of convergence at least R, and thus defines a function f, as implied by our notation. Indeed our inequalities show that f (k) (0) Bk! and consequently that c R k k B. R k If z < R, then c k z k B z k and this converges by comparison because z <. R To complete the proof, we must show that our subsequence converges uniformly to f on compact subsets of the interior of D. It is enough to show that it converges uniformly to f on z R < R. Pick a large N, and estimate the difference between f n and f carefully for the first N terms and then more broadly for the remaining terms, as follows: f n (z) f(z) N k! f n (k) (0) f (k) (0) z k + k=0 k=n+ 2B R k z k If ɛ > 0 is given and z R < R, we can make the last term smaller than ɛ 2 by choosing N large enough. After that we can make the first term smaller than ɛ 2 by choosing n large enough. QED. R 2.3 Compact Subsets of H(U) The main theorem of the previous section contains all the hard work. In this section, we generalize to an arbitrary U using abstract nonsense. Definition 9 A subset H of H(U) is locally uniformly bounded if for each compact K in U there is a uniform bound B such that the absolute value of every f H on K is at most B. Theorem 80 (Main Theorem; Montel) Let f n be a sequence of elements of H(U) and suppose this sequence is locally uniformly bounded. Then there is a subsequence of the original sequence which converges in H(U). Corollary 8 A subset H H(U) is compact in the topology of uniform convergence on compact subsets if and only if it is locally uniformly bounded and closed. Proof: We proved the key technical theorem for disks centered at the origin, but of course it remains true for disks with any center.

114 CHAPTER 2. TOPOLOGY ON H(U) 3 Choose a countable number of compact disks D i U of radius R i such that the corresponding open disks D i of radius R i 2 cover U. Our sequence is locally uniformly bounded on D i. So by the technical lemma, we can find a subsequence of {f n } which converges uniformly on the closure of D. By the same lemma, we can find a subsequence of this subsequence which converges uniformly on the closure of D2. Etc. By the diagonal argument, we can find a subsequence of the original sequence which is eventually a subsequence of each of these subsequences and so converges uniformly on the closure of each D i. A finite number of D i cover any compact subset of U, so the final subsequence converges uniformly on each such compact set. Proof of corollary: Since our H is a metric space, compactness of H follows from the main theorem. But we need to prove the converse, and for that it is enough to show that a compact subset of H must be locally uniformly bounded. This is easy. If H is compact but not locally uniformly bounded, then it is not uniformly bounded on some compact K. Find f n H and z n K with f n (z n ) > n. Suppose a subsequence f ni f uniformly on K. This f is continuous on the compact K and hence bounded. Since f ni f uniformly on K, the f ni are bounded on K, a contradiction.

115 Chapter 3 The Riemann Mapping Theorem 3. The Riemann Mapping Theorem In his PhD thesis, Riemann proved the following amazing theorem: Theorem 82 (Riemann Mapping Theorem) Let U be a simply-connected domain which omits at least one point of the plane. Then there is a one-to-one, onto holomorphic map f : U D to the open unit disk D about the origin. If z 0 U, we can choose f so f(z 0 ) = 0. If g is a second such map, there is a constant θ such that g(z) = e iθ f(z). Remark: The condition U C is essential, for if f : C D, then f is a bounded entire function and hence constant. Remark: Simply connected sets can be very wild and have extremely bad boundaries. See pictures below. So it is amazing that a continuous map exists, let alone a conformal map preserving angles. Even more amazing is uniqueness. A topological proof that a continuous homeomorphism exists would likely involve arbitrary choices, perhaps even uncountably many. This proof involves essentially no choices. Remark: Riemann s original theorem, from 85, had different hypotheses and a defective proof. It took over 50 years, until around 905, to reach the astonishing proof below. See the historical remarks after the proof. 4

116 CHAPTER 3. RIEMANN MAPPING THEOREM 5 Figure 3.: Simply Connected Set with Complicated Boundary Remark: In the above example, an infinite comb of vertical lines has been removed from a square, with x-coordinates n. The remaining set is simply connected, and can be mapped to the unit disk with z 0 mapping to the origin. It is unclear whether boundary points have a well-defined limit. Proof, Part : We first show that U can be mapped into the unit disk. The main problem is that U need not be bounded. Since U is not everything, we can translate it so 0 U. Since U is simply connected, we can define a branch of z on the translated U. If λ and λ are non-zero complex numbers, at most one can be in the range of z on U, because if λ = z and λ = z 2, then squaring gives z = z 2.. But z 0 is a non-zero number in the range of z; by the inverse function theorem a whole neighborhood V of z 0 is in the range of z. So no values of V are in the range. If v V, the map f(z) = z v is bounded on U and one-to-one. So the image is conformally equivalent to U and bounded. It is now easy to map this image into the unit disk by a one-to-one map. Proof, Part 2: Since the automorphism group of the unit disk is transitive, we can compose with an automorphism and get a map from U into the unit disk taking z 0 to zero. From

117 CHAPTER 3. RIEMANN MAPPING THEOREM 6 now on, replace U by its image in the disk and assume z 0 = 0. We are going to consider a family of maps f : U D that are one-to-one and fix 0. The first of these maps is the identity, and we will require that all subsequent maps satisfy f (0). Proof, Part 3: We now come to the heart of the proof. If f is a one-to-one map from U into the disk fixing 0, we will define a positive real number m(f) which measures how close f(u) is to filling the disk. We ll prove that if f is not onto, we can define a new map g with bigger m(g). Then we ll apply a compactness argument to find a map maximizing m. The obvious definition of m might be the area of f(u), but this definition doesn t work because the U below would have the maximal possible value and yet not be onto D. Figure 3.2: m(f) = area fails The stroke of genius, by I believe Koebe, is to select m(f) = df dz (0) Proof, Part 4: Lemma 7 Suppose f is not onto. Then there is a one-to-one map g : U D such that g(0) = 0 and g (0) and m(g) > m(f). Indeed, suppose w f(u) and define

118 CHAPTER 3. RIEMANN MAPPING THEOREM 7 g(z) = z w z w z z w z w f(z) Reading from right to left, the first map is f, which we want to improve. The second map is an automorphism of the disk which maps w to zero. Since w is not in the image of f, the composition of these two maps is never zero, so we can define a branch of the square root on this composition. The third map is this square root. We have already proved that the square root is one-to-one, so the full composition of the last three maps is a one-to-one map from U into D. This map sends the origin to w. The final map is an automorphism of the disk sending w back to the origin. So the entire composition is a one-to-one map g from U into the disk sending the origin to the origin. We next compute m(g) = dg. dz (0) For this we need d z a ( z a) (z a)( a) = dz z a ( z a) 2 = a 2 ( z a) 2 The required derivative is w 2 ( z w) 2 w 2 z w w 2 ( z w) 2 f (0) 0 = w 2 ( w 2 ) 2 2 w ( w 2 ) f (0) = + w 2 w f (0) The absolute value of this expression is larger than f (0) because Indeed, if 0 < a <, then +a2 2a >. + w 2 w > Proof, Part 5: So much for the concrete part of the proof. Let H H(U) be the set of all one-to-one holomorphic maps U D such that f(0) = 0 and f (0). Since each element of H is bounded by, Montel s theorem implies that H is compact provided it is closed. Suppose f n f where f n H. Then f is holomorphic on U and f(0) = 0 and f (0). We claim that f is one-to-one. This surprising result follows from the following lemma, since f (0) implies that f is not contant. Lemma 8 Suppose f n f uniformly on compact subsets of a domain U. If each f n is one-to-one, then either f is one-to-one or else it is constant.

119 CHAPTER 3. RIEMANN MAPPING THEOREM 8 Proof: Suppose not, so f(z ) = f(z 2 ) where z z 2. Replace U by U = {z 2 }, f n (z) by f n (z) f n (z 2 ), and f(z) by f(z) f(z 2 ). Then f n is never zero and f n f, but f has a zero. If f is not constant, this zero is isolated, so we can find a small counterclockwise circle γ around the zero with no other zeros on or inside the circle. Then f n fn f f uniformly on the circle, and so f n(ζ) 2πi γ f n (ζ) dζ f (ζ) 2πi γ f(ζ) dζ By the argument principle, this integral counts zeros inside γ, so the left side is zero and the right side is at least one, a contradiction. 3.2 History of the Riemann Mapping Theorem In Riemann s original formulation, the open domain U has a boundary and the map is to carry U to the interior of a disk and the boundary of U to the boundary circle. Riemann claimed the map was unique once we fix the image of one internal point z 0 and one boundary point b 0. There is no discussion of possible constraints on the boundary. Our earlier illustration shows that the boundary of general U can be very complicated. Later authors restricted to piecewise differentiable boundaries, or simple closed curves which only cross themselves at the endpoints. Alfors wrote in 953 that Riemann wrote almost cryptic messages to the future and his statement of the theorem has a form that would defy any attempt at proof, even with modern methods. In some sense, Riemann s proof relied on a boundary value problem for harmonic functions on a domain. Here are sketchy details, some in the form of exercises. Exercise 2 When discussing the Cauchy-Riemann equations, we wrote f = f +if 2. Here we adopt the alternate and common notation f = u + iv. Using the Cauchy-Riemann equations, show that if f is holomorphic then u is harmonic, that is 2 u x u y 2 = 0 Exercise 3 Suppose we know u and want to find v so u + iv is holomorphic. Cauchy-Riemann equations, we must solve v x = u y v y = u x By the

120 CHAPTER 3. RIEMANN MAPPING THEOREM 9 By advanced calculus, there is an integrability condition which must hold before these equations can be solved. Show that the integrability condition is exactly the statement that u is harmonic. Exercise 4 If u is harmonic on a simply-connected region U, show that v exists on U making u + iv holomorphic. Remark: We want a conformal map f : U D taking z 0 to zero. If this map is one-to-one, then z 0 must be a first order zero and we can form f(z) z z 0 a nowhere zero function on U. Since U is simply connected, this function equals e g for some g = u + iv. So we can assume that f = (z z 0 )e g(z) = e ln z z 0 +i arg(z z 0 ) e u+iv On the boundary of our U we want f to have absolute value one, so we must choose a harmonic u inside U with boundary values ln z z 0. Riemann was interested in physics as well as mathematics, and particularly interested in the theory of electricity and magnetism. In physics, harmonic functions describe steadystate conditions when a system has settled into a state independent of time. For instance, if we have a region with an insulated boundary and spray charges on the boundary, the potential inside is given by a harmonic function with boundary values given by the charges. Thus in physics, it is natural to assume that u exists. Riemann also knew of the Dirichlet principle. If u represents the potential energy function inside a region, the total energy is given by [ ( u ) 2 ( ) ] u 2 + dxdy x y R In a steady state, the potential energy should minimize this expression, and the Euler- Lagrange equations for this minimal energy state are exactly the condition that u be harmonic. So to obtain u, we minimize this expression among all possible u satisfying the given boundary conditions. Let us concentrate on this much of the proof, which produces u. We could then obtain v on the interior, but would still need to show that it also has appropriate boundary conditions. There is an interesting paper on the history of the Riemann Mapping theorem by Jeremy Grey, Rendiconti Del Circolo Mathematico Di Palermo, Serie II, Supplemento N. 34 (994). (Also available on the internet.) From this paper, we learn that Riemann s actual proof was more complicated than this sketch. However, Riemann s proof was shown to be invalid in 87 by his student Prym. Around the same time, Weierstrauss found an example of a calculus of variations problem for which the analogue of the Dirichlet principle failed, and

121 CHAPTER 3. RIEMANN MAPPING THEOREM 20 thus managed to convince mathematicians that an approach via the Dirichlet principle and harmonic functions was suspect. The next period from 887 to 90 was one of extensive work on the mapping theorem by Harnack, Hilbert, Osgood, Cartheodory, and Koebe. Koebe was a particularly important contributor. In 887, Harnack published a book which dramatically extended the theory of harmonic functions and solved the Dirichlet problem on many domains. The last pages of this book show how the theory can be used to prove the mapping theorem. Hilbert also worked on justifying the use of the Dirichlet principle for harmonic functions. Building on these results, the American mathematician Osgood may have been the first person to give a rigorous proof of the Riemann mapping theorem, between 900 and 902. He cleanly separated the case of mapping the interior from discussions of the boundary, and rigorously proved our form of the mapping theorem. He also proved that if the boundary is a continuously differentiable simple closed curve, then the map of the interior extends continuously to the boundary, and called attention to the possibility that this result might be true for an arbitrary continuous closed boundary curve which does not cross itself. Around the same time, Jordan proved the Jordan curve theorem: if ϕ : [0, ] C is continuous and ϕ(0) = ϕ() and otherwise ϕ is one-to-one, then ϕ separates the plane into two connected regions, one bounded and one unbounded. His proof was criticized, and then improved by Veblen, Schoenflies, and Brouwer in 909. Working with these results, Osgood proved that such a Jordan curve can have positive area. This made clear the difficulty of extending the mapping theorem to Jordan boundaries, let along more general boundaries. Caratheodory was encouraged by Klein and Hilbert to work on the mapping theorem. In papers published in 92 and 93, he replaced Riemann s uniqueness result that f is determined by f(z 0 ) and f(b 0 ) where b 0 is a boundary point, with the assertion that f is determined by f(z 0 and the argument of f (z 0 ) and then very cleanly proved the Riemann mapping theorem on the interior of the domain. According to Gray, this is the first purely function theoretic proof, with no appeal to Harnack s work, replacing that approach by the use of Schwarz s lemma. In a second paper, Caratheodory proved that the conformal map extends to a homeomorphism of the boundaries if and only if the boundary is a simple Jordan curve in the sense described above. Gray ends his history of the Riemann mapping theorem by writing: That the boundary of a domain could be as strange as Osgood and Caratheodory had discovered was a powerful stimulus to precision in these matters. It is reasonable to claim that in proving the Riemann mapping theorem,... mathematicians came to unite geometric function theory

122 CHAPTER 3. RIEMANN MAPPING THEOREM 2 with topology, to the mutual advantage of both. Indeed, a consequence of Caratheodory s 93 paper is that if ϕ : S C is a simple closed curve, then there is a homeomorphism of the plane mapping the image of ϕ to the unit circle, and mapping the bounded component of the complement conformally to the disk and the unbounded component conformally to the exterior of the disk. The Jordan curve theorem generalizes this result by proving that if S n R n is a one-to-one continuous map, then the complement of the image has two components, one bounded and one unbounded. Around 920, Alexander proved that in three dimensions, the bounded component is homeomorphic to the open ball B 3 R 3. His paper was accepted. As Alexander waited for publication, he came to have more and more doubts, and eventually he found the Alexander horned sphere, a counterexample which made him immortal. So the authors of papers on Riemann s result were working on a cliff. In the midst of the above development, Poincare transformed the subject. In the early part of the 9th century, a central topic was the study of meromorphic functions on a torus, or what is the same thing, doubly periodic meromorphic functions on the plane. We later devote a chapter to that theory. Now Poincare attempted to generalize this subject by replacing the plane by the unit disk. This is a natural object to consider because the disk has a large number of automorphisms, which can be used to map regions to other regions in the same manner that translations map rectangular regions in the plane to other rectangular regions. Poincare had to find subgroups Γ of the automorphism group which are discrete. A fundamental region for such a group is a domain U in the disk such that no two elements of U are equivalent under Γ each element of the entire disk is equivalent under Γ to at least one point in the closure of U. In that case, the elements γ(u) cover the disk in the same way that rectangles cover the plane. Memomorphic functions on the disk which are invariant under Γ correspond to meromorphic functions on the Riemann surface D/Γ. Applying the conformal equivalence of the disk and the upper half plane, Poincare s task was equivalent to finding discrete subgroups of SL(2, R). A few such subgroups like SL(2, Z) were known from the earlier theory, and when finding other examples proved difficult, Poincare at first tried to prove that the known examples were the only possibilities. At this point we can follow Poincare exactly, because he later wrote a book called The Psychology of Mathematical Invention about his next steps. Poincare was about to go on vacation, and the night before he drank strong coffee. Then he found that he could not sleep. The next morning, just as he was about to step on the bus, it occurred to him that

123 CHAPTER 3. RIEMANN MAPPING THEOREM 22 the unit disk behaves like non-euclidean geometry. Poincare reports that he then relaxed on vacation, and wrote up the results when he returned to Paris. The connection to non-euclidean geometry allowed Poincare to define regions directly using the kinds of arguments in Euclid, but suitably modified for the non-euclidean case. This quickly led to discrete groups whose quotient spaces D/Γ were compact surfaces of genus g 2. Poincare made the audacious guess that all such surfaces arise in this way. The proof turned on the following generalization of the Riemann mapping theorem: Theorem 83 (The Uniformization Theorem) Every simply connected Riemann surface is conformally equivalent to S 2, D, or C Much of the work on the mapping theory from 900 on was directed toward proving this generalization. Alfors once wrote that the Riemann mapping theorem is one of the most important theorems of complex analysis and the uniformation theorem is perhaps the single most important theorem in the whole theory of analytic functions of one variable. 3.3 A Short Sketch of Consequences of the Uniformization Theorem 3.3. Manifolds Having talked of Riemann surfaces several times, we finally give a definition. Let us start with ordinary continuous surfaces. Definition 20 An n-dimensional manifold M is a topological Hausdorff space which is second countable, together with a covering of M by coordinate neighborhoods (U, ϕ, V). Here U M is an open set in M, ϕ is a homeomorphism from U to V, and V R n is an open set in R n. The essential idea here is that small pieces of M look topologically like small pieces of Euclidean space. The Hausdorff and second countable conditions eliminate bizarre examples like the long line and will not be mentioned again. Note that U varies over only some of the open sets; to give M, it suffices to give a covering of M by enough such coordinate systems. Remark: If (U, ϕ, V) is a coordinate neighborhood, then each point of V has coordinates (x, x 2,..., x n ) and we can use these to describe points on the manifold via the map ϕ. Usually, we ignore all the fancy words and just say suppose we have coordinates (x,..., x n ) defined on U. Definition 2 If U, ϕ, V and U 2, ϕ 2, V 2 are coordinate systems, we get a map from the

124 CHAPTER 3. RIEMANN MAPPING THEOREM 23 plane to the plane as in the following picture. It is called the glueing map. Such maps explain how the various coordinate systems glue together. Remark: This map is only defined over the common intersection. In more detail ϕ 2 (ϕ ) : ϕ (U U 2 ) ϕ 2 (U U 2 ) Example: Recall that we made the Riemann sphere into a manifold by covering it with two coordinates U = C z z C and U 2 = C {0} { } C, so the glue map is /z. Example: Usually we avoid this fancy terminology. The point is that some points of M can belong to intersections with two coordinates: (x,..., x n ) and (y,..., y n ). The glue map explains how to convert from x s to y s or conversely. This is done by giving conversion functions y i = y i (x,..., x n ). Figure 3.3: Glue map ϕ 2 (ϕ ) If all of the glue maps are C maps from the plane to the plane, we say that M is a C manifold. If the dimension of M is two and all of the glue maps are holomorphic, we say that M is a Riemann surface. Moise and Munkries proved that any manifold of dimension less than or equal to three has a differentiable structure, which is unique up to differomorphism. (This is dramatically false in dimensions four and above.) So we can assume that surfaces are C Riemannian Manifolds If we have a C manifolds, we can give it an additional structure allowing us to compute distances and angles; such a structure is known as a Riemannian metric. Suppose we have

125 CHAPTER 3. RIEMANN MAPPING THEOREM 24 local coordinates near a point p, allowing us to introduce coordinates (x,..., x n ). Then if γ (t) and γ 2 (t) are differentiable curves through p, we can compute dγ dt = X and dγ 2 dt = Y and get two tangent vectors at p. A Riemannian structure on M assigns an inner product < X, Y > to any two such vectors defined at p. This makes it possible for us to compute angles between vectors, and lengths of vectors. By linear algebra, < X, Y >= g ij X i Y j where g ij is a symmetric positive definite matrix, so equivalently a Riemannian structure is given by functions g ij on each coordinate system. Riemannian structures were defined on surfaces by Gauss in his great work on differential geometry. The extension to higher dimensional manifolds was sketched by Riemann in his required Habilitationsschrift (inaugural lecture) for the PhD (Gauss was on his committee). For surfaces, Gauss defined a quantity called the Gaussian curvature and explained how to compute this function given the g ij. A necessary and sufficient condition condition that a change of coordinates converts g ij to the Euclidean form δ ij is that k = 0 nearby. In other cases, k serves to distinguish various geometries defined by the g ij. Riemann showed how to extend this k to the Riemannian curvature tensor, an analogous object in higher dimensions. The quantity k is usually a function, which makes it difficult to handle. In special cases, however, it is a constant. By magnifying, this constant is either, 0, or -. In these cases, Gauss proved that k completely determines the local geometry. The cases in question are k = : standard sphere k = 0: standard flat plane k = : non-euclidean geometry of Bolyai and Lobachevsky Remark: After Riemann s work, this division into three cases was extended to all dimensions, and the crucial hypotheses became clearer. An isometry of a Riemannian manifold is a one-to-one and onto map which preserves inner products, and so distances and angles. A fairly easy theorem asserts that if p and q are points and v,..., v n is an orthonormal basis of tangent vectors at p and w,..., w n is a similar basis of tangent vectors at q, then there is at most one isometry mapping p to q and the v i to the w i. So an isometry is determined by the image of p and the corresponding rotation about this image. For most Riemannian M, isometries are rare. Remark: But suppose M is a Riemannian manifold such that all possible isometries occur. The great theorem of the subject asserts that there are only four possibilities up to

126 CHAPTER 3. RIEMANN MAPPING THEOREM 25 magnification (provided the dimension of M is at least two). M can be the sphere S n or the plane R n, or the generalized non-euclidean geometry H n. Or finally, M can be S n with opposite points identified. This space is called projective space. Since it is locally isometric to S n, many people ignore it and assert that there are only three very symmetric Riemannian manifolds. Every C manifold can be given a Riemannian metric. The argument employs partitians of unity, that is, functions which are identically near a point p, but then rapidly vanish and are identically zero off the coordinate system. A partition of unity is a collection ϕ α of such functions such that ϕ α =. To get a global g ij, we define g ij in each coordinate systems, and then form ϕ α g ij,α. We mention this technique because it would never be used in complex analysis, since the identity theorem would imply that each ϕ α is identically zero Riemann Surfaces Given g ij on a surface, coordinates exist such that g ij = δ ij only when the curvature is zero. However, Gauss and later many others proved that isothermal coordinates always exist, where g ij = λ(x, y)δ ij. This means that every metric is Euclidean up to magnifications which vary from place to place. In particular, any metric is conformally equivalent to the standard metric of the plane. Suppose, then, that M is an orientable Riemannian surface. Cover M by isothermal coordinates. The glueing maps are than conformal and hence holomorphic, so M becomes a Riemann surface Covering Surface In topology, there is a beautiful theory of covering spaces. It won t be described here, but will be used (only in this chapter). Incidentally, after a twenty year gap, Boubaki has published a new volume, about covering spaces and the fundamental group. Suppose M is a Riemann surface. Then we can form its universal covering space, M. Since M is covered by coordinates with holomorphic glueing maps, this also holds for M, which becomes a Riemann surface. Moreover, the deck transformation group maps must be holomorphic.

127 CHAPTER 3. RIEMANN MAPPING THEOREM The Uniformization Theorem Applying the uniformization theorem, we conclude that M is either the Riemann sphere, the complex plane, or the unit disk. Moreover, the deck transformation group Γ is a discrete subgroup of the automorphism group of M without fixed points, so if γ id, then γ(z) z. A short calculation shows that every linear fractional transformation has fixed points. So if the universal covering space is the sphere, then the original surface is also the sphere. We conclude that Theorem 84 The Riemann sphere has a unique Riemann surface structure. Consequently, any Riemannian surface homeomorphic to a sphere has a conformally equivalent second metric equal to the standard metric on a sphere. Remark: If the universal cover is the plane, then every deck transformation has the form ϕ(z) = az + b. Such maps have fixed points unless a =. Consequently the deck transformation group must be a discrete subgroup formed only of translations. It is easy to conclude that the only possibilities are z z + l where l runs through all elements of a lattice of rank 0,, or 2. Theorem 85 The only surfaces with universal cover C are C, cylinders, and tori. If a Riemannian manifold is homeomorphic to one of these, it has a conformally equivalent second metric which is flat. Remark: Cylinders are conformally equivalent to annuli. Remark: In all other cases, the universal cover is a disk. Astonishingly, the entire automorphism group of a disk preserves the non-euclidean metric. Consequently, and astonishingly Theorem 86 Every Riemannian surface except the sphere, the complex plane, cylinders, and tori can be given a conformally equivalent metric locally equivalent to non-euclidean geometry. Remark: This result is false in higher dimensions. But Thurston conjectured that it is essentially correct in dimension three. The three geometries of the theorem must be replaced by eight geometries, but all except the non-euclidean case can essentially be analyzed completely. Moreover, a three-dimensional manifold must be cut into pieces and then the result says that each piece can be given one of these eight geometrical structures. Remark: This difficult conjecture implies the Poincare conjecture. Like the uniformization theorem, it is proved using difficult theorems in analysis about differential equations. The complete proof has now been given by Hamilton and Perelman, with expositions from a number of Chinese mathematicians.

128 Chapter 4 The Theorems of Mittag-Leffler and Weierstrass 4. The Mittag-Leffler Theorem Definition 22 If z 0 is a pole of f, the principal part of f at z 0 is the sum of negative terms in the Laurent expansion of f about z 0 : c k (z z 0 ) k + c k+ (z z 0 ) k c z z 0 More generally, a principal part P(z) at z 0 is a polynomial in and without constant term. z z 0 of degree at least one Theorem 87 (Mittag-Leffler) Let z, z 2,... be a sequence in the complex plane with no limit points. Let P k (z) be a principal part at z k, for each k. Then there is a meromorphic function f(z) on C with poles precisely at the z i, and principal part P k at z k. Proof: If one of the z i is zero, set it aside and let Q i = P i. For each remaining k, expand P k in a power series c i z i about the origin. This series converges for 0 < z k, and converges uniformly for 0 z k 2. Let Q k be j Q k = P k c i z i i=0 where j has been chosen so Q k 2 k on 0 z k 2. Let f(z) = Q k (z) 27

129 CHAPTER 4. THE THEOREMS OF MITTAG-LEFFLER AND WEIERSTRASS 28 There are many ways to see that this converges appropriately and solves the problem. For example, choose a large open disk U of radius R centered about the origin. The closed disk of radius 2R is compact, so only finitely many z k are inside this disk. Choose M so k > M implies z k is not in the larger disk. Then f(z) = M k= Q k + g(z) where g(z) is the sum of the remaining terms. Each Q k in this remaining sum has absolute value less than on U, so g(z) is a uniform limit of holomorphic functions on U and thus holomorphic. 2 k QED. Theorem 88 (Mittag-Leffler) Let U be a domain in the plane. Let z, z 2,... be a sequence in U with no limit points in U. Let P k (z) be a principal part at z k, for each k. Then there is a meromorphic function f(z) on U with poles precisely at the z i, and principal part P k at z k. Proof: The z k can now have limit points on the boundary of U; limit points can also occur at infinity. For each k, let Consider two sets of indices d k = inf w U z k w A = { k d k z k } B = { k d k z k > } In some sense, A contains indices of z k which go to the boundary faster than they go to infinity, and B contains indices of z k which go to infinity faster than they go to the boundary. We will define f(z) = f (z) + f 2 (z) where f handles poles for indices in A and f 2 handles poles for indices in B. Our first claim is that the previous version of the Mittag-Leffler theorem handles B. Indeed, if the corresponding z k had a finite limit point in the plane, this limit cannot be in U, and so for an appropriate subsequence, d k 0. Since d k z k >, we conclude that z k are unbounded, a contradiction. So it suffices to construct f for A. If k A, a compactness argument finds a b k in the boundary of U with d k = z k b k. Expand the principal part P k about b k in a Laurent series which converges in the annulus z b k > d k. See the picture near the top of the next page.

130 CHAPTER 4. THE THEOREMS OF MITTAG-LEFFLER AND WEIERSTRASS 29 For example, suppose P = z and b k = 0. The expansion we need is P = z = z ( z ) = z k 0 ( z ) k = k ( ) k z Figure 4.: Modifying Principal Parts to Force Convergence In general, this Laurent series will have only terms of negative powers and converge uniformly to P as long as we stay away from the inner boundary. For each k, choose j such that j Q k = P k c i (z b k ) i is smaller than 2 k on z b k > 2d k, and let i= f (z) = Q k (z) This f has the correct principal parts and the modifying terms z b k have singularities on the boundary but remain holomorphic on U. It suffices to show that on each compact subset of U, Q k converges uniformly after we remove the first few terms. So let K U be compact. If for all but finitely many k we have z b k > 2d k for z K, then for all but finitely many k we have Q m < and the sum over these remaining terms 2 k converges uniformly on K. We are in trouble if for infinitely many k we can find w k K with w k b k 2d k. From now on, restrict attention to these k. In this bad case, one possibility is that d k 0. This would imply that K is arbitrarily close to the boundary of U, and this is impossible. If the d k to not converge to zero, we can restrict attention to an infinite subset bounded below by, say, B. So 0 < B d k and B z k d k z k, and so z k B. It follows that

131 CHAPTER 4. THE THEOREMS OF MITTAG-LEFFLER AND WEIERSTRASS 30 these z k have a convergent subsequence. The limit must not belong to U, so d k 0, a contradiction. Remark: Mittag-Leffler was a wonderful Swedish mathematician with many accomplishments, including the above theorems which made him immortal. In the United States, a standard mathematical rumor asserts that there is no Nobel prize in mathematics because Mittag-Leffler had an affair with Nobel s wife. Curiously, this story is not part of standard lore in Europe. One difficulty with the rumor is that Nobel never married. Remark: The Mittag-Leffler theorem is the complex analysis version of partial fraction reduction as taught in integration theory. 4.2 Infinite Products We now prepare for Weierstrauss theorem, an extension of the high school task of factoring polynomials into linear terms. Remark: We want to define infinite products g(z) = Πg k (z). We d like various theorems, true for sums, to remain true for products: The convergence criterion should be related to uniform convergence on compact subsets. It should automatically follow that the product of holomorphic functions is again holomorphic It should be possible to rearrange convergent products without changing the result There should be a simple formula for the derivative of a product. We want to get all of this for free without doing any work! In addition, we want The infinite product is zero if and only if one of the factors is zero. The order of Πg k (z) at z 0 should be the sum of the orders of the g i at z 0. In a convergent product, g i (z) as i. Remark: The naive definition does not have these properties. If the c k are complex and we define Πc k = lim n Πn k= c i then it is easy to find examples where the product is zero but none of the terms are zero, and examples where the product converges and yet the terms do not approach one.

132 CHAPTER 4. THE THEOREMS OF MITTAG-LEFFLER AND WEIERSTRASS 3 Remark: Searching for the correct definition, we recall that logarithms convert products to sums. Perhaps the theory reduces to the known theory of sums using Πe Log(g i) = e Log(g i ) The problem is that we cannot take the logarithm of functions with zeros, and the logarithm is multi-valued. On the other hand, when g i is close to one, the principal branch of the logarithm is well-defined and this formula makes sense. Remark: This leads to the formal definition: Definition 23 If c i are complex numbers such that c i, there is an N such that for i > N we have R c i > 0. Then the standard branch of the logarithm Log(c i ) is defined. If i>n Log(c i) converges absolutely, we say that the infinite product converges and define to be (c c 2... c N ) e i=n Log(c i ) Lemma 9 This product does not depend on the index when we switch to taking logarithms. Proof: Notice that e Log(c i) = c i. Any remaining words of proof are just fluff. 4.3 A Convergence Criterion for Products The previous definition requires us to determine if Log(c i ) converges. This can be difficult. Since c i is close to, write c i = + d i. The crucial lemma of the theory of infinite products asserts that our sum converges absolutely if and only if d i converges absolutely. Testing this condition is much easier. Begin with the following series, which converges if z < : Therefore if z < 2, we have and Log( + z) z Log( + z) = z z2 2 + z Log( + z) z + z 2 + z = z z < 2 z ( z ) 2 z 2 2 z z ( )... = 2 z Notice that if c i and c i = + d i, then eventually d i < 2 and after that, Log(c i ) converges if and only if d i converges. We say the infinite product of the f i converges on U if on each compact subset K of U there is an N such that

133 CHAPTER 4. THE THEOREMS OF MITTAG-LEFFLER AND WEIERSTRASS Infinite Products of Holomorphic Functions Definition 24 Let f i (z) be holomorphic on a domain U. Write f i (z) = + g i (z). We say the infinite product of the f i converges on U if for each compact K U, g i converges uniformly on K. In that case, the infinite product is given by fi = (f (z) f 2 (z)... f N (z)) e i>n Log(f i (z)) In this formula, i > N implies f i (z) < 2. Remark: The sum of logarithms converges uniformly on compact K, so this function is holomorphic on U. Because the sum converges absolutely, the product can be rearranged at will. It is clear that the order of the product at z 0 is the (finite) sum of orders of factors at z 0. Remark: Finally, we discuss discuss the derivative of such a product. Definition 25 If f(z) is holomorphic, the logarithmic derivative of f is f (z) f(z) Remark: Notice that this is the derivative of the logarithm of f, provided this logarithm is defined. Notice also that the log of a product is the sum of the logs, and thus the logarithmic derivative of a product is the sum of the logarithmic derivatives of the terms. Theorem 89 If each f i (z) is holomorphic on U, and f ( z) exists, we have d dz fi fi = f i f i Proof: This is trivially true for finite sums. It also holds for the term e i>n Log(f i (z)) since the exponent can be differentiated term by term by a previous result. 4.5 The Theorem of Weierstrauss Weierstrass theorem does for zeros what Mittag Leffler s theorem does for poles:

134 CHAPTER 4. THE THEOREMS OF MITTAG-LEFFLER AND WEIERSTRASS 33 Theorem 90 (Weierstrass) Let z, z 2,... be a sequence in the complex plane with no limit points. Let k, k 2,... be a related sequence of positive integers. Then there is an entire function f(z) with zeros precisely at the z i, each of order k i. Remark: The proof is similar to the proof of the Mittag-Leffler theorem. The functions f i (z) = (z z i ) k i have zeros of the required orders at z i. We modify these functions to obtain convergence, and then take their infinite product. An infinite product converges if the terms are closer and closer to, so the modified functions we seek should have a zero at z i and yet be very close to. Concentrate on the case when z i =. We propose the modified function E(z) = ( z)e Log( z) because it has a zero at and yet is identically one. This logarithm does not exist at, so we form an approximation. Since z = + z + z2 +..., integration gives Log( z) = z + z2 2 + z Replacing the logarithm with a finite approximation, we are led to consider Definition 26 ( ) z+ E n (z) = ( z)e z zn n Lemma 0 If z, then E n (z) z n+ Proof: The derivative of E n (z) is ( ) z+ ( z)e z zn ( n + z z n ) z+ e( z zn n ) ( ) = z n z+ e z zn n It is clear that the exponential term expands in a power series whose coefficients are nonnegative real numbers. So and de n dz = zn a z n+ a 2 z n+2... z n+2 z n+3 E n (z) = zn+ n + a n + 2 a 2 n Thus E n (z) z n+ = b 0 + b z + b 2 z for b i 0. If z, E n (z) z n+ b i z i b i = E n() = and the inequality follows.

135 CHAPTER 4. THE THEOREMS OF MITTAG-LEFFLER AND WEIERSTRASS 34 Figure 4.2: E 2 (z) and E 3 (z) Figure 4.3: E 4 (z) Proof of Weierstrass Theorem Renumber the given points to w, w 2, w 3,... so each z i is repeated k i times, and 0 < w w 2 w (If z = 0 is supposed to be a zero of multiplicity k, replace f by z k f(z) at the ed of the argument.) Define f(z) = E i ( z w i ) Incidentally, this is a conservative choice making convergence easy to prove; in many situations each term could be E 2. By the previous section, this converges provided E i ( z w i )

136 CHAPTER 4. THE THEOREMS OF MITTAG-LEFFLER AND WEIERSTRASS 35 converges uniformly on compact subsets K of the plane. Since the w i have no limit points, eventually z w i 2 on K. Applying the inequality at the start of this section, eventually ( ) z E i z i+ w i i+ 2 and we have the desired convergence. QED. Theorem 9 (Weierstrass) Let U be a domain in the plane. Let z, z 2,... be a sequence in U with no limit points in U. Let k, k 2,... be a related sequence of positive integers. Then there is a holomorphic function f(z) on U with zeros precisely at the z i, each of order k i. Proof: The proof follows the outline of the similar Mittag-Leffler proof. For each z i we find a closest boundary point b i and distance d i = z i b i. Separate indices into A = { k b i z i } and B = { k b i z i > }. Find functions f and f 2 with zeros in A and B, respectively, and define f(z) = f (z) + f 2 (z). The z k with indices in B have no finite limit points, so the previous argument gives f 2 (z). Rename the indices in A so each occurs k i times. We want to handle these indices as in the Mittag-Leffler proof, perhaps using the same proof pictured below. w i Figure 4.4: Modifying E n (z) to Force Convergence The previous proof used expansion of Laurent series in an annulus. Our job is to find an analogue for E n (z). We have accumulated many tools, and one that comes to mind is linear fractional transformations, which move points on the sphere. Perhaps we could employ ( ) Az + B E n Cz + D

137 CHAPTER 4. THE THEOREMS OF MITTAG-LEFFLER AND WEIERSTRASS 36 A linear fractional transformation is determined by the images of three points. If our choice mapped b i to infinity, then the modified E n would have a singularity at b i, which is not in U. If our choice mapped z i to, then the modified E n would have a zero at z i. Finally, if our choice mapped to 0, then points far away, i.e., not close to b i, would be mapped to near zero where E n is almost one. Thus we are led to form ( ) zi b i E n z b i Below, for example, is this function where we require a zero at 2.0, and the closest boundary point is 3.0. This function belows up in a complicated way at 3, which is not in U, but has our required zero at 2. Moreover is close to away from the region containing these two points. According to the crucial lemma about E n, if Figure 4.5: E 4 ( zi b i z b i ); z i = 2; b i = 3 z i b i z b i, then ( ) E zi b i n z b i z i b i n+ z b i Apply this when z i b i z b i 2, or equivalently z b i 2 z i b i, then ( ) ( E zi b i n z b i 2 ) n+

138 CHAPTER 4. THE THEOREMS OF MITTAG-LEFFLER AND WEIERSTRASS 37 So define f (z) = ( ) zi b i E i z b i As in the proof of the general Mittag-Leffler theorem, if K is compact in U and indices belong to A, then for all but finitely many i we have z b i > 2d i and so ( ) zi b i E n z b i converges uniformly. Corollary 92 If U is a domain, there is a holomorphic function f(z) on U which cannot be extended to be holomorphic on any larger connected open set. Proof: Find a sequence z n U without limit points in U but such that every boundary point of U is a limit point. Let f(z) have first order orders at each z n and no other zeros. Any large open set would contain limit points of the zeros of f, so f would be identically zero on the larger set. Remark: There is a difficult theory of functions of several complex variables, invented mostly in the 20th century. One reason the theory is difficult is that the previous result is no longer true. Instead, open sets U V exist with the property that all functions holomorphic in U can be extended to be holomorphic in V.

139 Chapter 5 Classical Sum and Product Formulas 5. Beautiful Identities We get beautiful formulas when we apply the techniques of the Mittag-Leffler and Weierstrass theorems to poles or zeros with a symmetrical pattern, and try to adjust the Principal Part or E(z) as little as possible for convergence. Since the trigonometric functions have periodic zeros, our formulas often sum to trigonometric expressions. Theorem 93 π sin πz = z + ( ( ) n z n + ) n n 0 π 2 cos πz sin 2 πz = z 2 + ( ) n (z n) 2 = n 0 n ( ) n (z n) 2 π cos πz sin πz = z + n 0 ( (z n + ) n π 2 sin 2 πz = z 2 + (z n) 2 = n 0 n (z n) 2 Remark: In the first formulas, the principal parts are z n and up to signs, which (z n) 2 alternate. A brief check shows that the second formula is obtained by differentiating the first. The principal parts of the third and fourth formulas are always z n and (z n) 2 38

140 CHAPTER 5. CLASSICAL SUM AND PRODUCT FORMULAS 39 without sign alternation, and again the final formula is the derivative of the third one. By differentiating again and again, we obtain similar formulas with principal parts, (z n) k with or without sign alternations. After multiplying by constants and adding, we can get any principal part, repeated with or without sign alternations. Next we check that the right hand sides converge. It suffices to study the first and third formulas because convergence of the other sums follows by differentiation. In the first and third formulas, we could not simply write z n because the resulting sums would not converge. For instance, if we set z = 0 and ignore the singular term z, we get n which diverges. Consequently, we modified the terms using Mittag-Leffler s procedure. If we expand z n about the origin, we get ( ) z n = n z n = n ( z n) k = k z k n k+ According to Mittag-Leffler, we get convergence by subtracting a finite number of these terms, and our formula suggests that it is enough to subtract n. This works because z n + n = z (z n)n If K is a compact subset of the plane and z K, then the numerator of this fraction is bounded by, say, B, and for n large enough, we have (z n)n ( n 2 n). So the series converges because < and n 2 z (z n)n 4B n 2 Notice that no Mittag-Leffler modification is required for series with higher powers of z n. Next we claim that the left and right sides of the first two equations satisfy g(z+) = g(z) and consequently are periodic of period 2; similarly we claim that the left and right sides of the third and fourth equations are periodic of period. This is completely trivial for the left sides by ordinary trigonometry. It is also trivial for the right sides of the second and fourth formula because all integers are treated the same up to sign in these formulas. A slight argument is required for the right sides in the first and third formulas, and we ll give it for the first formula and sketch it for the third. Let g(z) be the sum of the right side of the first formula. Since the second formula is the derivative of the first, and the second formula is periodic up to sign, we have d dz g(z + ) = d dz g(z) and so d dz (g(z + ) + g(z)) = 0. Consequently, g(z + ) + g(z) is constant. This

141 CHAPTER 5. CLASSICAL SUM AND PRODUCT FORMULAS 40 constant is g(/2) + g( /2). However, g(z) is an odd function, so the sum of these two terms is zero. Indeed, g(z) = z + ( ( ) n z n + n + z + n ) = n z + 2z (z 2 n 2 ) n>0 n>0 The argument for the third formula is similar; we want to prove that g(z + ) = g(z), but their derivatives are equal, so g(z + ) g(z) is constant. In particular, set z = 2. Then g ( ) ( ( 2 g 2) is this constant. As before, g is odd, so the constant is 2g 2). However, g(z) = z + ( z n + ) z + n n>0 so g ( ) = 2 + ( 2 2 2n + 2 ) [ = 2 + ( + 2n 2n + ) ] 2n + n>0 n>0 and the value inside the square brackets is ( + + ) ( ) ( ) +... = 0 7 Next we claim that the left sides of our four formulas have the same principal parts as the right sides, and consequently the difference between the two sides is an entire function. This is proved using the same sort of calculation in all four cases, so we only consider the first case. In this first case, g(z + ) = g(z), so it suffices to check the principal parts at the origin. But sin z = z z3 3! + z5 5! +... and thus sin πz = πz π3 z = πz ( π2 z 2 ) ! 3! and so π sin πz = ( π2 z 2 ) +... = z 3! z The final step of the proof of the second and fourth identities proceeds as follows. Let z = x + iy where 0 x and y. We prove that the left and right sides of our identities go to zero. It follows from this and periodicity that the difference between right and left sides is bounded, hence constant. But since the above limit is zero, this constant must be zero.

142 CHAPTER 5. CLASSICAL SUM AND PRODUCT FORMULAS 4 We carry this out for the second formula. Notice that cos πz = eiπz + e iπz 2 = eiπx e πy + e iπx e πy 2 sin πz = eiπz e iπz = eiπx e πy e iπx e πy 2i 2i We consider the case y, letting the reader handle the case y. Then all terms in these expressions go to zero except the last term, which is in absolute value eπy 2 in both cases. But in our expression the cos πz term is in the numerator, and the sin πz term is squared in the denominator. So the quotient goes to zero. This argument works for the first, second, and fourth formulas. Next look at the right side. The second and fourth formulas are the same up to an irrelevant ( ) n. So consider. If z is anywhere in the strip 0 x, then z n n (z n) 2 2 for all but finitely many n. For the remaining terms, (z n) 4. 2 n 2 Suppose we want to make the right side less than ɛ by choosing y large enough. Choose N so n >N 4 < ɛ n 2 2. Only finitely many terms remain to be considered, but their sum can also be made less than ɛ 2 because lim z = 0. (z n) 2 This completes the proof for the second and fourth formulas. However, the argument just given does not work for the first and third formulas. (Try it; the individual terms go to zero, but getting an initial bound on infinitely many terms doesn t work.) So a trick is required in the first and third cases. But the derivative of the difference between the left and right sides of the first formula is the difference between the left and right sides of the second formula, hence zero. It follows that the difference between the left and right sides of the first formula is constant. Similarly the difference between the left and right sides of the third formula is constant. To show that both of these constants are zero, it suffices to evaluate the first and third formulas at specific points. A good choice is z = 2. We do both calculations at once. Recall that the two formulas can be rewritten as π sin πz = z + ( ( ) n z n + ) z + n n 0 π cos πz sin πz = z + n 0 ( z n + ) z + n

143 CHAPTER 5. CLASSICAL SUM AND PRODUCT FORMULAS 42 Substituting z = 2 gives or or π = 2 + ( 2 ( ) n 2n + 2 ) + 2n n 0 0 = 2 + n 0 ( 2 2n + 2 ) + 2n π 2 = + ( ( ) n 2n + ) 2n + n 0 0 = + ( 2n + ) 2n + n 0 ( π 2 = + ) ( ) ( ) ( 0 = + + ) ( ) ( ) or 0 = ( ) + π 4 = ( 3 ) ( ) + 5 ( 7 7 ) +... Both of these formulas are true; the first was one of the first results deduced by Leibniz during the invention of the calculus. This finishes the argument for all four formulas. Example Substituting z = 2 in the fourth formula gives π 2 = 4 (2n ) 2 and so If we set then π 2 8 = L = L π2 8 = = ( ) = L 4

144 CHAPTER 5. CLASSICAL SUM AND PRODUCT FORMULAS 43 and so and 3L 4 = π2 8 L = π2 6 = Remark: Below is a plot of the third formula. In this case, the limit of the absolute value of the function high above the x-axis is not zero; indeed it is π. Figure 5.: π cos πz sin πz 5.2 A Product Formula Theorem 94 sin πz = πz n 0 ( z ) ) e z n = πz ( z2 n n 2 n>0 Proof: From the proof of Weierstrauss theorem, recall that ( ) z+ E n (z) = ( z)e z zn n